Techniques of antidifferentiation - AP Calculus AB
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A projectile is shot up from a platform 
above the ground with a velocity of 
. Assume that the only force acting on the projectile is gravity that produces a downward acceleration of 
. Find the velocity as a function of 
.
A projectile is shot up from a platform
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$a=\frac{d^2s$$}{dt^2$}$=\frac{dv}{dt}$=-9.8 with initial conditions v(0)=150
Separate velocity variables and solve.
v=-9.8t+C
Plug in intial conditions
v=-9.8t+150
$a=\frac{d^2s$$}{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
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?
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
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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.
The bullet will be at a height of 500 ft on the way up and on the way down.
We use the position equation
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
The position of a particle as a function of time is given below:
At what values of 
does the particle change direction?
The position of a particle as a function of time is given below:
At what values of
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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:
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 
.
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,
If we set
The possible points where
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
Thus,
Lastly, we will evaluate the velocity at a value of
The sign of the velocity has switched back to positive, so the particle does indeed change direction at
The answer is
A jogger leaves City 
at
. His subsequent position, in feet, is given by the function:
,
where 
is the time in minutes.
Find the velocity of the jogger at 15 minutes.
A jogger leaves City
where
Find the velocity of the jogger at 15 minutes.
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To find velocity, one has to use the first derivative of 
:
.
Note the units have to be ft/min.
To find velocity, one has to use the first derivative of
Note the units have to be ft/min.
The speed of a car traveling on the highway is given by the following function of time:
Note that
What does this mean?
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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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.
The function
Use a change of variable (aka a u-substitution) to evaluate the integral,
Use a change of variable (aka a u-substitution) to evaluate the integral,
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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 
. Now if we look at the original integral we can rewrite in terms of 
Now proceed with the integration with respect to 
.
Now write the result in terms of 
using equation (1), we conclude,
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,
Now differentiate with respect to
Looking at equation (2), we can solve for
Now proceed with the integration with respect to
Now write the result in terms of
Solve the following integral using substitution:
Solve the following integral using substitution:
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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 
in place of 
into the original integral and simplify.
The 
in the denominator cancels out the remaining 
in the integral, leaving behind a 
. We can pull the 
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. 
2.
3. 
4. 
5. 
6. 
7. 
8. 
9. 
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
The
The specific steps are as follows:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Solve the following integral using substitution:
Solve the following integral using substitution:
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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 
in place of 
into the original integral and simplify. The 
in the denominator cancels out the remaining 
in the integral, leaving behind a 
. We can pull the 
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. 
2. 
=
3. 
4.
5. 
6. 
7. 
8. 
To solve the integral, we have to simplify it by using a variable
1.
2.
3.
4.
5.
6.
7.
8.
Solve the following integral using substitution:
Solve the following integral using substitution:
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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 
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. 
2. 
3. 
4. 
5. 
6. 
7. 
To solve the integral, we have to simplify it by using a variable
1.
2.
3.
4.
5.
6.
7.
Solve the following integral using substitution:
Solve the following integral using substitution:
Tap to see back →
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 
in place of 
into the original integral and simplify. The 
in the denominator cancels out the remaining 
in the integral, leaving behind a 
. We can pull the 
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. 
2. 
3. 
4. 
5. 
6. 
7. 
To solve the integral, we have to simplify it by using a variable
1.
2.
3.
4.
5.
6.
7.
Solve the following integral using substitution:
Solve the following integral using substitution:
Tap to see back →
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 
in place of 
into the original integral and simplify. The 
in the denominator cancels out the remaining 
in the integral, leaving behind a 
. We can pull the 
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. 
2. 
3. 
4. 
5. 
6. 
7. 
To solve the integral, we have to simplify it by using a variable
1.
2.
3.
4.
5.
6.
7.
Solve the following integral using substitution:
Solve the following integral using substitution:
Tap to see back →
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 
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. 
2. 
3. 
4. 
5. 
6. 
7. 
To solve the integral, we have to simplify it by using a variable
1.
2.
3.
4.
5.
6.
7.
Solve the following integral using substitution:
Solve the following integral using substitution:
Tap to see back →
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 
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. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
To solve the integral, we have to simplify it by using a variable
1.
2.
3.
4.
5.
6.
7.
8.
Solve the following integral using substitution:
Solve the following integral using substitution:
Tap to see back →
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 
in place of 
into the original integral and simplify. The 
in the denominator cancels out the remaining 
in the integral, leaving behind a 
. We can pull the 
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. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
To solve the integral, we have to simplify it by using a variable
1.
2.
3.
4.
5.
6.
7.
8.
Solve:
Solve:
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To integrate, we must make the following substitution:
Rewriting the integral in terms of u and integrating, we get
The following rule was used for integration:
Replacing u with our original x term, we get
To integrate, we must make the following substitution:
Rewriting the integral in terms of u and integrating, we get
The following rule was used for integration:
Replacing u with our original x term, we get
Use u-subtitution to fine
Use u-subtitution to fine
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Let 
Then 
Now we can subtitute
Now we substitute back
Let
Then
Now we can subtitute
Now we substitute back
Evaluate 
Evaluate
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We can use substitution for this integral.
Let 
,
then 
.
Multiplying this last equation by 
, we get 
.
Now we can make our substitutions
. Start
. Swap out 
with 
, and 
with 
. Make sure you also plug the bounds on the integral into 
for 
to get the new bounds.
. Factor out the 
.
. Integrate (absolute value signs are not needed since 
.)
. Evaluate
.
We can use substitution for this integral.
Let
then
Multiplying this last equation by
Now we can make our substitutions
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This is a u-substitution integral. We need to substitute the new function, which is modifying our base function (the exponential).
, but instead of that, our problem is 
. We can solve this integral by completing the substitution.
Now, we can replace everything in our integrand and rewrite in terms of our new variables:
.
Remember to plug your variable back in and include the integration constant since we have an indefinite integral.
This is a u-substitution integral. We need to substitute the new function, which is modifying our base function (the exponential).
Now, we can replace everything in our integrand and rewrite in terms of our new variables:
Remember to plug your variable back in and include the integration constant since we have an indefinite integral.
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This is a u-substitution integral. We need to choose the following substitutions:
Now, we can replace our original problem with our new variables:
In the last step, we need to plug in our original function and add the integration constant.
This is a u-substitution integral. We need to choose the following substitutions:
Now, we can replace our original problem with our new variables:
In the last step, we need to plug in our original function and add the integration constant.
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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!
Now, let us rewrite our original equation in terms of our new variable!
.
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!
Now, let us rewrite our original equation in terms of our new variable!