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 .

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?

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?

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.

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?

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:

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

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

Solve the following integral using substitution:

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

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

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

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

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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Question

Solve the following integral using substitution:

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

Answer

To solve the integral, we have to simplify it by using a variable u to substitute for a variable of x.

For this problem, we will let u replace the expression .

Next, we must take the derivative of u. Its derivative is .

Next, solve this equation for dx so that we may replace it in the integral.

Plug in place of and $\frac{du}{4x^3}$ in place of into the original integral and simplify.

The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{4}$ . We can pull the $\frac{1}{4}$ out front of the integral. Next, take the anti-derivative of the integrand and replace u with the original expression, adding the constant to the answer.

The specific steps are as follows:

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

2.

3.

4. $dx=\frac{du}{4x^3}$

5. $\int{x^3u^5(\frac{du}{4x^3})}$

6. $\frac{1}{4}\int{u^5}du$

7. $\frac{1}{4}(\frac{1}{6}u^6)$

8. $\frac{1}{4}(\frac{1}{6}(2+x^4)^6$

9. $\frac{1}{24}(2+x^4)^6+c$

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Question

Solve the following integral using substitution:

$\int{x^2\sqrt{x^3+1}}dx$

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 u. 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. $\int{x^2\sqrt{x^3+1}}dx$

2. =

3.

4. $\int{x^2\sqrt{u}\frac{du}{3x^2}}$

5. $\frac{1}{3}\int{\sqrt{u}du}$

6. $\frac{1}{3}(\frac{2}{3}u^{\frac{3}{2}})+c$

7. $\frac{2}{9}u^{\frac{3}{2}}+c$

8. $\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

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Question

Solve the following integral using substitution:

$\int{cos^3xsinxdx}$

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 $\frac{du}{-sinx}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral, leaving behind a . 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. $\int{cos^3xsinxdx}$

2.

3.

4. $\int{u^3sinx(\frac{du}{-sinx})}$

5. $\int-u^3du$

6. $\frac{-1}{4}u^4+c$

7. $\frac{-cos^4x}{4}+c$

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Question

Solve the following integral using substitution:

$\int{e^xcos(e^x)dx}$

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 $\frac{du}{e^x}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in 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. $\int{e^xcos(e^x)dx}$

2.

3.

4. $\int{e^xcosu(\frac{du}{e^x})}$

5. $\int{cosudu}$

6.

7.

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Question

Solve the following integral using substitution:

$\int{\frac{sin(lnx)}{x}dx}$

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 $\frac{1}{x}$ . Next, solve this equation for so that we may replace it in the integral. Plug in place of and in place of into the original integral and simplify. The in the denominator cancels out the remaining in 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. $\int{\frac{sin(lnx)}{x}dx}$

2.

3. $du=(\frac{1}{x}dx)$

4.

5. $\int{\frac{sinu}{x}dux}$

6. $\int{sinudu}$

7.

8.

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Question

Solve the following integral using substitution:

$\int{(x^2+e^{3x})(x^3+e^{3x})^{\frac{6}{7}}dx}$

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 $x^3+e^{3x}$ . Next, we must take the derivative of . Its derivative is $3x^2+e^{3x}$ . Next, solve this equation for so that we may replace it in the integral. Plug in place of $x^3+e^{3x}$ and $\frac{du}{3x^2+e^{3x}}$ in place of into the original integral and simplify. The $x^2+e^{3x}$ in the denominator cancels out the remaining $x^2+e^{3x}$ 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. $\int{(x^2+e^{3x})(x^3+e^{3x})^{\frac{6}{7}}dx}$

2. $u=x^3+e^{3x}$

3. $du=(3x^2+3e^{3x})dx$

4. $\int(x^2+e^{3x})u^{\frac{6}{7}(\frac{du}{x^2+e^{3x}})}$

5. $\frac{1}{3}\int{u^{\frac{6}{7}}}du$

6. $\frac{1}{3}(\frac{7}{13})u^{\frac{13}{7}}$

7. $\frac{7}{39}u^{\frac{13}{7}}+c$

8. $\frac{7}{39}(x^3+e^{3x})^{\frac{13}{7}}+c$

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Question

Use u-subtitution to fine

$\int \sin^2(x)\cos (x)dx$

Answer

Let

Then

Now we can subtitute

$\int sin^2(x) \cos(x) dx=\int u^2du= \frac{1}{3}u^3$

Now we substitute back

$\frac{1}{3}u^3=\frac{1}{3} \sin^3(x)$

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Question

Evaluate $\int_0^1 \frac{6x}{1+x^2} dx$

Answer

We can use substitution for this integral.

Let ,

then .

Multiplying this last equation by , we get .

Now we can make our substitutions

$\int_0^1 \frac{6x}{1+x^2} dx$ . Start

$=\int_1^2 \frac{3}{u}du$ . Swap out with , and with . Make sure you also plug the bounds on the integral into for to get the new bounds.

$=3\int_1^2 \frac{1}{u}du$ . Factor out the .

$=3 \ln(u)|_1^2$ . Integrate (absolute value signs are not needed since $1\le u \le2$ .)

$=3(\ln(2)-\ln(1))$ . Evaluate

$=3\ln2 -0 = 3\ln2$ .

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Question

$\int e^{4x-2}dx=?$

Answer

This is a u-substitution integral. We need to substitute the new function, which is modifying our base function (the exponential).

$\int e^xdx=e^x+C$ , but instead of that, our problem is . We can solve this integral by completing the substitution.

$u=4x-2.; du=4dx. ;dx=\frac{du}{4}.$

Now, we can replace everything in our integrand and rewrite in terms of our new variables:

$\int e^{4x-2}dx=\int e^u\frac{du}{4}=\frac{1}{4}\int e^udu.$

$=\frac{1}{4}e^u+C=\frac{1}{4}e^{4x-2}+C$ .

Remember to plug your variable back in and include the integration constant since we have an indefinite integral.

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Question

$\int 2cos(2x)dx=?$

Answer

This is a u-substitution integral. We need to choose the following substitutions:

Now, we can replace our original problem with our new variables:

$\int2cos(2x)dx=\int cos(2x)2dx=\int cos(u)du$

In the last step, we need to plug in our original function and add the integration constant.

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Question

$\int \sqrt{3+4x}dx=?$

Answer

This is a hidden u-substitution problem! Because we have a function under our square root, we cannot just simply integrate it. Therefore, we need to choose the function under the square root as our substitution variable!

$u=3+4x.; du=4dx.; dx=\frac{du}{4}.$

Now, let us rewrite our original equation in terms of our new variable!

$\int \sqrt{3+4x}dx=\int \sqrt{u}\frac{du}{4}=\frac{1}{4}\int \sqrt udu.$

$=\frac{1}{4}\int u^{1/2}=\frac{1}{4}\frac{u^{3/2}}{3/2}+C=\frac{1}{4}*\frac{2}{3}u^{3/2}+C$

$=\frac{2}{12}u^{3/2}+C=\frac{1}{6}({3+4x)}^{3/2}+C$ .

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Question

$\int \frac{2lnx}{x}dx=?$

Answer

This is a hidden u-substitution problem! Remember, to use substitution, we need to have an integral where a function and its derivative live inside. If you look closely, you will see we have just that!

$u=lnx.; du=\frac{1}{x}dx.$

Now, rewrite the integral, and integrate:

$\int \frac{2lnx}{x}==2\int lnx*\frac{1}{x}dx=2\int u du=2\frac{u^2}{2}+C$

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