Integrals - AP Calculus AB

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

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Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Tap to reveal answer

Answer

has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\ \ \end{matrix} \left|\begin{matrix} 0 \ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \ \ \end{matrix}\right| \begin{matrix} \pi \ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

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Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

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Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

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