Use of the Fundamental Theorem to evaluate definite integrals - AP Calculus AB

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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 →

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.

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Question

Evaluate

$\int_{-1}^{e-2}\frac{1}{x+2}$

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Answer

We will use the Fundamental Theorem of Calculus

$\int_{a}^{b}f(x)dx=F(b)-F(a)$

and the rule

$\int\frac{1}{x+a}dx=\ln(x+a)$

First we find the anti derivative

$\int_{-1}^{e-2}\frac{1}{x+2}=\ln(x+2)|_{-1}^{e-2}$

And then we evaluate, (upper minus lower)

$\ln(x+2)|_{-1}^{e-2}=\ln(e-2+2)-\ln(-1+2)=\ln(e)-\ln(1)$

(Remembering your logarithm rules)

$\ln(e)-\ln(1)=1-0=1$

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Question

Using the Fundamental Theorem of Calculus solve the integral.

$\int_{1}^{3}e^xdx$

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Answer

To solve the integral using the Fundamental Theorem, we must first take the anti-derivative of the function. The anti-derivative of is . Since the limits of integration are 1 and 3, we must evaluate the anti-derivative at these two values.

denotes the anti-derivative.

When we do this,

and .

The next step is to find the difference between the values at each limit of integration, because the Fundamental Theorem states

$\int_{a}^{b}f(x)dx=F(b)-F(a)$ .

Thus, we subtract to get a final answer of .

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Question

Using the Fundamental Theorem of Calculus and simplify completely solve the integral.

$\int_{3}^{6}\frac{dx}{x}$

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Answer

To solve the integral, we first have to know that the fundamental theorem of calculus is

$\int_{a}^{b}f(x)dx=F(b)-F(a)$ .

Since denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 3 and 6.

To find the anti-derivative, we have to know that in the integral, $\frac{dx}{x}$ is the same as $\frac{1}{x}dx$ .

The anti-derivative of the function $\frac{1}{x}$ is , so we must evaluate .

According to rules of logarithms, when subtracting two logs is the same as taking the log of a fraction of those two values:

$ln\frac{6}{3}$ .

Then, we can simplify to a final answer of

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Question

Solve $\int_{0}^{3}(x^3-6x)dx$ using the Fundamental Theorem of Calculus.

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Answer

To solve the integral, we first have to know that the fundamental theorem of calculus is

$\int_{a}^{b}f(x)dx=F(b)-F(a)$ .

Since denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 0 and 3.

The anti-derivative of the function is $\frac{1}{4}x^4-3x^2$ , so we must evaluate .

When we plug 3 into the anti-derivative, the solution is $\frac{-27}{4}$ , and when we plug 0 into the anti-derivative, the solution is 0.

To find the final answer, we must take the difference of these two solutions, so the final answer is $\frac{-27}{4}$ .

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Question

Solve $\int_{0}^{2}(2x^3-6x+\frac{3}{x^2+1})$ using the Fundamental Theorem of Calculus.

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Answer

To solve the integral, we first have to know that the fundamental theorem of calculus is

$\int_{a}^{b}f(x)dx=F(b)-F(a)$ .

Since denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 0 and 2.

The anti-derivative of the function

$2x^3-6x+\frac{3}{x^2+1}$

is

$\frac{1}{2}x^4-3x^2+3tan^{-1}x$ ,

so we must evaluate .

When we plug 3 into the anti-derivative, the solution is $-4+3tan^{-1}(2)$ , and when we plug 0 into the anti-derivative, the solution is 0.

To find the final answer, we must take the difference of these two solutions, so the final answer is $-4+3tan^{-1}(2)$ .

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Question

Use the fundamental theorem of Calculus to evaluate the definite integral

$\int_{\pi/2}^{\pi}(2cos(x)+1)dx$

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Answer

Here we use the fundamental theorem of Calculus: $\int_{a}^{b}f(x)dx=F(b)-F(a)$

$F(x)=\int(2cos(x)+1)dx=2sin(x)+x$

Here we do not worry about adding a constant c because we are evaluating a definite integral.

$F(b)=F(\pi)=2sin(\pi)+\pi=0+\pi=\pi$

$F(a)=F(\frac{\pi}{2})=2sin(\frac{\pi}{2})+\frac{\pi}{2}= 2+\frac{\pi}{2}$

$F(b)-F(a)=F(\pi)-F(\frac{\pi}{2})=\pi-(2+\frac{\pi}{2})=\frac{\pi}{2}-2$

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Question

Evaluate $\int_1^x\frac{2}{t^2}dt$ .

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Answer

We can integrate this without too much trouble

$\int_1^x\frac{2}{t^2}dt$ . Start

$=\int_1^x2t^{-2}dt$ . Rewrite the power

$=[-2t^{-1}]_1^x$ . Integrate

$=(-2x^{-1})- (-2)$ . Evaluate

$=\frac{2(x-1)}{2}$ . Simplify

Note that we were not asked to evaluate $\frac{d}{dx}\int_1^x\frac{2}{t^2}dt$ , so you should not attempt to use part one of the Fundamental Theorem of Calculus. This would give us the incorrect answer of $\frac{2}{x^2}$ .

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Question

Evaluate the indefinite integral:

$\int_0^{\frac{\pi}{4}} \tan(x)\sec^2(x)dx$

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Answer

$\int_0^{\frac{\pi}{4}} \tan(x)\sec^2(x)dx$

First compute the indefinite integral:

$\int \tan(x)\sec^2(x)dx$

Note that the $\sec^2(x)$ is the derivative of $\tan(x)$ . So proceed by defining a new variable:

$u = \tan(x) : : : : :\rightarrow : : : : :du=\sec^2(x)dx$

Now the the integral can be written in terms of

$\int \tan(x)\sec^2(x)dx=\int udu =\frac{1}{2}u^2 +C=\frac{1}{2}\tan^2(x)+C$

Therefore:

$\int \tan(x)\sec^2(x)dx=\frac{1}{2}\tan^2(x)+C$

When we go to compute the indefinite integral the constant of integration will be ignored since it will be subtracted out when we evaluate.

We can precede by either going back to the original variable and evaluate over the original limits of integration, or we can find new limits of integration corresponding to the new variable . Let's look at both equivalent methods:

Solution 1)

$\int_0^{\frac{\pi}{4}}\tan(x)\sec^2(x)dx=\left[ \frac{1}{2}\tan^2(x)\right ]_{x=0}^{x=\frac{\pi}{4}}: : =: : : : \frac{1}{2}\tan^2\left(\frac{\pi}{4}\right) -\frac{1}{2}\tan^2\left(0\right)$

$\tan(0)=0$ so the last term vanishes. The first term reduces to $\frac{1}{2}$ since the tangent function is equal to .

$\int_o^{\frac{\pi}{4}}\tan(x)\sec^2(x)dx=\frac{1}{2}$

Solution 2)

We could have also solved without converting back to the original variable. Instead, we could just change the limits of integration. Use the definition assigned to the variable , which was $u = \tan(x)$ and then use this to find which value takes on when (lower limit) and when $x =\frac{\pi}{4}$ (upper limit).

$x = 0: : : : : : :\rightarrow: : : : :u = \tan(0)=0$

$x=\frac{\pi}{4}: : : : :\rightarrow : : : : :u=tan\left(\frac{\pi}{4}\right )=1$

$\int_0^1 udu =\left[\frac{1}{2}u^2 \right ]_{u=0}^{u=1} =\frac{1}{2}$

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Question

$\frac{d}{dx}\int_{0}^{x^2} sin(t)dt=?$

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Answer

Derivatives and anti derivatives annihilate each other. Therefore, the derivative of an anti derivative is simply the function in the integral with the limits substituted in multiplied by the derivative of each limit. Also, be careful that the units will match the outermost units (which comes from the derivative).

$\frac{d}{dx}\int_{0}^{x^2} sin(t)dt= sin(x^2)\frac{d}{dx}(x^{2})-sin(0)\frac{d}{dx}(0)'$

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