Techniques of antidifferentiation - AP Calculus AB

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Question

A projectile is shot up from a platform $5\ m$ above the ground with a velocity of $150\ m/s$ . Assume that the only force acting on the projectile is gravity that produces a downward acceleration of $9.8\ m/s^{2}$ . Find the velocity as a function of .

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Answer

$a=\frac{d^2s}{dt^2}=\frac{dv}{dt}=-9.8$ with initial conditions $v(0)=150$

Separate velocity variables and solve.

$v=-9.8t+C$

Plug in intial conditions

$v=-9.8t+150$

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Question

A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after seconds. What is its velocity 500 ft into the air?

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Answer

The bullet will be at a height of 500 ft on the way up and on the way down.

We use the position equation to solve for how long it will take to reach a height = 500. and seconds.

We then plug that into the velocity equation, which is the derivative of the position function. .

We can see that plugging in the value of yields and yields . The positive and negative values of velocity indicates the up and down direction of travel.

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Question

The position of a particle as a function of time is given below:

$x(t)=\frac{t^{3}}{3}-3t^{2}+8t$

At what values of does the particle change direction?

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Answer

In order to find the point at which the particle changes direction, we must determine whenever the velocity of the particle changes sign (from positive to negative, or from negative to positive).

We will need to have the function of the particle's velocity before we can determine where the velocity changes sign. Because the velocity is the derivative of position with respect to time, we can write the function for velocity, , as follows:

$v(t)=x'(t)=\frac{d}{dx}[\frac{t^{3}}{3}-3t^{2}+8t]=\frac{3t^{2}}{3}-6t+8=t^{2}-6t+8$

If we set , then we can determine the points where it can change sign.

The possible points where will change signs occur at . However, we need to check to make sure.

First, we can try a value less than 2, such as 1, and then a value between 2 and 4, such as 3. We will evaluate at and and see if the sign of the velocity changes.

Thus, is indeed a point where the velocity changes sign (from positive to negative). This means that the particle does in fact change direction at .

Lastly, we will evaluate the velocity at a value of larger than 4, such as 5.

The sign of the velocity has switched back to positive, so the particle does indeed change direction at .

The answer is and .

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Question

A jogger leaves City at. His subsequent position, in feet, is given by the function:

$x(t) = 5t^{2} - 3t + 2$ ,

where is the time in minutes.

Find the velocity of the jogger at 15 minutes.

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Answer

To find velocity, one has to use the first derivative of :

.

Note the units have to be ft/min.

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Question

The speed of a car traveling on the highway is given by the following function of time:

Note that

What does this mean?

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Answer

The function gives you the car's speed at time . Therefore, the fact that means that the car's speed is at time . This is equivalent to saying that the car is not moving at time . We have to take the derivative of to make claims about the acceleration.

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Question

Solve the following integral using substitution:

$\int{xsin(x^2)dx}$

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Answer

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression . Next, we must take the derivative of . Its derivative is . Next, solve this equation for so that we may replace it in the integral. Plug in place of and $\frac{du}{2x}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{2}$ . We can pull the $\frac{1}{2}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{xsin(x^2)dx}$

2.

3.

4. $\int{x(sinu)\frac{du}{2x}}$

5. $\frac{1}{2}\int{sinudu}$

6. $\frac{-1}{2}cosu+c$

7. $\frac{-1}{2}cos(x^2)+c$

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Question

Find the antiderivative of the following.

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Answer

Follow the following formula to find the antiderivatives of exponential functions:

$\int x^n=\frac{x^n^+1}{n+1}$

Thus, the antiderivative of is $\frac{x^3}{3}$ .

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Question

Find the antiderivative of the following.

$\cos (x)$

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Answer

$\cos (x)$ is the derivative of $\sin (x)$ . Thus, the antiderivative of $\cos (x)$ is $\sin (x)$ .

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Question

Find the antiderivative of the following.

$\sec^2(x)$

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Answer

$\sec^2(x)$ is the derivative of $\tan (x)$ . Thus, the antiderivative of $\sec^2(x)$ is $\tan (x)$ .

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Question

Solve the following integral using substitution:

$\int{x^2e^{x^3}dx}$

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Answer

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression . Next, we must take the derivative of . Its derivative is . Next, solve this equation for so that we may replace it in the integral. Plug in place of and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{3}$ . We can pull the $\frac{1}{3}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{x^2e^{x^3}dx}$

2.

3.

4. $\int{x^2e^u(\frac{du}{3x^2})}$

5. $\frac{1}{3}\int{e^udu}$

6. $\frac{1}{3}e^u+c$

7. $\frac{1}{3}e^{x^3}+c$

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Question

Evaluate the integral

$\int (4x^5+3x^2+1 )dx$

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Answer

To evaluate the integral, we use the rules for integration which tell us

$\int x^ndx=\frac{x^{n+1}}{(n+1)}+C$

Applying to the integral from the problem statement, we get

$\frac{4}{6}x^6+x^3+x+C$

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Question

Evaluate the following integral

$\int (\sin(x)+\frac{x}{4})dx$

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Answer

To solve the problem, we apply the fact that anti-derivative of $\sin(x)=-\cos(x)$ and that $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Taking the anti-derivative of each part independently, we get

$\int \sin(x)dx+\int \frac{x}{4}dx$

Finally, our answer is

$-\cos(x)+\frac{x^2}{8}+C$

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Question

Evaluate the following integral

$\int (3x^3+2x-5 )dx$

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Answer

To evaluate the integral, we use the definition

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

$\int (3x^3+2x-5 )dx=\frac{3}{4}x^4+x^2-5x+C$

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Question

Evaluate the following integral

$\int (\sin{x}+\sec^2{x)}dx$

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Answer

To evaluate the integral, we use the fact that the antiderivative of $\sin{x}$ is $-\cos{x}$ (because $\frac{\mathrm{d} }{\mathrm{d} x}(-\cos{x})=\sin{x}$ ), and the antiderivative of $\sec^2{x}$ is $\tan{x}$ (because $\frac{\mathrm{d} }{\mathrm{d} x}(\tan{x})=\sec^2x$ ). Using this information, we determine that the integral is

$-\cos{x}+\tan{x}+C$

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Question

Calculate the following integral.

$\int4sin(t)-12cos(t)+3t^2dt$

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Answer

Calculate the following integral.

$\int4sin(t)-12cos(t)+3t^2dt$

To do this problem, we need to recall that integrals are also called antiderivatives. This means that we can calculate integrals by reversing our integration rules.

Thus, we can have the following rules.

$\int sin(t)dt = -cos(t)+c$

$\int cos(t)dt = sin(t)+c$

$\int t^ndt=\frac{t^{n+1}}{n+1}+c$

Using these rules, we can find our answer:

$\int4sin(t)-12cos(t)+3t^2dt$

Will become:

$-4cos(t)-12sin(t)+\frac{3t^3}{3}+c=-4cos(t)-12sin(t)+t^3+c$

And so our answer is:

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Question

Calculate the integral in the following expression:

$\int(e^2)^{\ln 3x}dx$

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Answer

Solving this integral depends on knowledge of exponent rules; mainly, that $(a^b)^c = a^{bc}$ . Using this, we can simplify the given expression.

$\int(e^2)^{\ln3x}dx = \int e^{2 \ln 3x}dx = \int (e^{\ln 3x})^2dx =\int (3x)^2dx =\int 9x^2dx$

From here, we just follow the power rule, raising the exponent and dividing by it.

$9\int x^2 dx = 3x^3 + C$

Giving us our final answer.

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Question

Evaluate the integral

$\int (4x^4+2x^2+3x )dx$

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Answer

To evaluate the integral, we use the following definition

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

$\int (4x^4+2x^2+3x )dx=\frac{4x^5}{5}+\frac{2x^3}{3}+\frac{3x^2}{2}+C$

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Question

Integrate:

$\int (x^2 +\frac{1}{2x})dx$

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Answer

To evaluate the integral, we can split it into two integrals:

$\int x^2 dx+ \frac{1}{2}\int \frac{dx}{x}$

After integrating, we get

$\frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |+C$

where a single constant of integration comes from the sum of the two integration constants from the two individual integrals, added together.

The rules used to integrate are

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int \frac{dx}{x}=\ln\left | x\right | +C$

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Question

Solve:

$\int \frac{dx}{12+9x^2}$

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Answer

The integral is equal to

$\frac{1}{2\sqrt{3}}\arctan(\frac{3x}{2\sqrt{3}})+C$

and was found using the following rule:

$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\arctan{\frac{x}{a}}+C$

where $a=\sqrt{12}=2\sqrt{3}$

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Question

Use a change of variable (aka a u-substitution) to evaluate the integral,

$\int x\sqrt{1+x^2}dx$

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Answer

$\int x\sqrt{1+x^2}dx$

Integrals such as this are seen very commonly in introductory calculus courses. It is often useful to look for patterns such as the fact that the polynomial under the radical in our example, , happens to be one order higher than the factor outside the radical, You know that if you take a derivative of a second order polynomial you will get a first order polynomial, so let's define the variable:

(1)

Now differentiate with respect to to write the differential for ,

(2)

Looking at equation (2), we can solve for , to obtain $xdx =\frac{1}{2}du$ . Now if we look at the original integral we can rewrite in terms of

$\int x\sqrt{1+x^2}dx = \int\sqrt{1+x^2}\underbrace{xdx} = \int\sqrt{u}\left(\frac{1}{2}du \right )$

Now proceed with the integration with respect to .

$\int\sqrt{u}\left(\frac{1}{2}du \right ) = \frac{1}{2}\int u^{\frac{1}{2}}du$

$= \frac{1}{2}\left[\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} \right ]+C$

$=\frac{1}{2} \left(\frac{2}{3}u^{\frac{3}{2}} \right )+C$

$=\frac{1}{2} \left(\frac{2}{3}u^{\frac{3}{2}} \right )+C$

$=\frac{1}{3}u^{\frac{3}{2}}+C$

Now write the result in terms of using equation (1), we conclude,

$\int x\sqrt{1+x^2}dx=\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C$

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