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AP Calculus AB : Integrals

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Example Question #1 : Integrals

Evaluate $\frac{d}{dx}\int_4^{x^2} e^{t^2}dt$ .

Possible Answers:

$2xe^{x^4}$

$x^2e^{x^2}$

$e^{x^2}$

Does not exist

$\frac{e^{t^2}}{t^2}$

Correct answer:

$2xe^{x^4}$

Explanation:

Even though an antideritvative of $e^{t^2}$ does not exist, we can still use the Fundamental Theorem of Calculus to "cancel out" the integral sign in this expression.

$\frac{d}{dx}\int_4^{x^2} e^{t^2}dt$ . Start

$= e^{(x^2)^2} \times (x^2)'$ . You can "cancel out" the integral sign with the derivative by making sure the lower bound of the integral is a constant, the upper bound is a differentiable function of , f(x) , and then substituting t = f(x) in the integrand. Lastly the Theorem states you must multiply your result by f'(x) (similar to the directions in using the chain rule).

$=2xe^{x^4}$ .

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Example Question #2 : Integrals

The graph of a function f(x) is drawn below. Select the best answers to the following:

Pbstm

What is the best interpretation of the function?

$F(x)=\int_a^xf(t)dt$

Which plot shows the derivative of the function F(x) ?

Possible Answers:

Wrong3q10

Question 10 correct answer

Wrngan2

Wrn4

Correct answer:

Question 10 correct answer

Explanation:

$F(x)=\int_a^xf(t)dt$

The function F(x) represents the area under the curve f(x) from x=a to some value of x>a .

Do not be confused by the use of in the integrand. The reason we use is because are writing the area as a function of , which requires that we treat the upper limit of integration as a variable . So we replace the independent variable of f(x) with a dummy index when we write down the integral. It does not change the fundamental behavior of the function F(x) or f(t) .

The graph of the derivative of F(x) is the same as the graph for f(x) . This follows directly from the Second Fundamental Theorem of Calculus.

If the function f(x) is continuous on an interval containing , then the function defined by:

$F(x)=\int_a^xf(t)dt$

has for its' derivative F'(x)=f(x) .

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Example Question #1 : Integrals

Evaluate

$\int_{-(4\pi+e^4)}^{4\pi+e^4}3\sin(2x)dx$

Possible Answers:

$\frac{-3\cos(8\pi+2e^4)}{2}+\frac{3\cos(-8\pi-2e^4)}{2}$

$\frac{-3\cos(8\pi+2e^4)}{2}$

$\frac{-3\cos(\pi+e^4)}{2}$

Correct answer:

Explanation:

Here we could use the Fundamental Theorem of Calculus to evaluate the definite integral; however, that might be difficult and messy.

Instead, we make a clever observation of the graph of

$3\sin(2x)$

Namely, that

$3\sin(2x)=-3\sin(-2x)$

This means that the values of the graph when comparing x and -x are equal but opposite. Then we can conclude that

$\int_{-(4\pi+e^4)}^{4\pi+e^4}3\sin(2x)dx=\int_{-(4\pi+e^4)}^{0}3\sin(2x)dx+\int_{0}^{4\pi+e^4}3\sin(2x)dx$

=-(SOMETHING)+(SOMETHING)=0

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Example Question #1 : Techniques Of Antidifferentiation

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 .

Possible Answers:

$v=-150+9.8$

$v=-5t+9.8$

$v=-9.8t+150$

$v=-5t+150$

$v=-9.8t+5$

Correct answer:

$v=-9.8t+150$

Explanation:

$a=\frac{d^2s}{dt^2}=\frac{dv}{dt}=-9.8$ 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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Example Question #1 : Integrals

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?

Possible Answers:

$158.1\ and\ -158.1$

$138.1\ and -138.1$

$-128.1\ and\ 128.1$

Correct answer:

$-128.1\ and\ 128.1$

Explanation:

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

We use the position equation (s(t)) to solve for how long it will take to reach a height = 500. t=2.87 and 10.88 seconds.

We then plug that into the velocity equation, which is the derivative of the position function. v=220-32t .

We can see that plugging in the value of t=10.88 yields v=-128.1 and t=2.87 yields v=128.1 . The positive and negative values of velocity indicates the up and down direction of travel.

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Example Question #1 : Techniques Of Antidifferentiation

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?

Possible Answers:

$t=3\ only$

$t=0\ only$

$t=2\ only$

$t=3\ and\ t=1$

$t=2\ and\ t=4$

Correct answer:

$t=2\ and\ t=4$

Explanation:

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, v(t) , as follows:

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

v(t)=t^2-6t+8

If we set v(t)=0 , then we can determine the points where it can change sign.

v(t)=t^2-6t+8=0

(t-4)(t-2)=0

t = 2, 4

The possible points where v(t) will change signs occur at t=2,4 . 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 v(t) at t=1 and t=3 and see if the sign of the velocity changes.

v(1)=1^2-6(1)+8=2

v(3)=3^2-6(3)+8=-1

Thus, t=2 is indeed a point where the velocity changes sign (from positive to negative). This means that the particle does in fact change direction at t=2 .

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

v(5)=5^2-6(5)+8=3

The sign of the velocity has switched back to positive, so the particle does indeed change direction at t=4 .

The answer is t=2 and t=4 .

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Example Question #1 : Integrals

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

v(t) = 5t^2+2t

Note that

v(0) =0

What does this mean?

Possible Answers:

The car is not decelerating at time t=0 .

The car's speed is constantly changing at time t=0 .

The car takes seconds to reach its maximum speed.

The car is not moving at time t=0 .

The car is not accelerating at time t=0 .

Correct answer:

The car is not moving at time t=0 .

Explanation:

The function v(t) gives you the car's speed at time . Therefore, the fact that v(0)=0 means that the car's speed is at time . This is equivalent to saying that the car is not moving at time t=0 . We have to take the derivative of to make claims about the acceleration.

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Example Question #8 : Integrals

A jogger leaves City at t=0 . 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.

Possible Answers:

$147\ ft/min$

$120\ ft/min$

$147\ ft$

$372\ ft/min$

$150\ ft/min$

Correct answer:

$147\ ft/min$

Explanation:

To find velocity, one has to use the first derivative of f(t) :

v(t) = x'(t) = 10t - 3 .

v(15) = 10(15) - 3 = 147 ft/min.

Note the units have to be ft/min.

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Example Question #1 : Integrals

Write the equation of a tangent line to the given function at the point.

y = ln(x²) at (e, 3)

Possible Answers:

y – 3 = (2/e)(x – e)

y – 3 = ln(e²)(x – e)

y = (2/e)(x – e)

y – 3 = (x – e)

y = (2/e)

Correct answer:

y – 3 = (2/e)(x – e)

Explanation:

To solve this, first find the derivative of the function (otherwise known as the slope).

y = ln(x²)

y' = (2x/(x²))

Then, to find the slope in respect to the given points (e, 3), plug in e.

y' = (2e)/(e²)

Simplify.

y'=(2/e)

The question asks to find the tangent line to the function at (e, 3), so use the point-slope formula and the points (e, 3).

y – 3 = (2/e)(x – e)

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Example Question #1 : Integrals

Find the equation of the tangent line at (1,-1) when

$y=\frac{x^{3}-2x(x^{1/2})}{x}$

Possible Answers:

y=x

y=x-2

y=x+1

y=1

y=-x-2

Correct answer:

y=x-2

Explanation:

The answer is

$y=\frac{x^{3}-2x(x^{1/2})}{x}$ let's go ahead and cancel out the 's. This will simplify things.

$y=x^{2}-2(x^{1/2})$

$y'=2x-\frac{1}{(x^{1/2})}$

$y'(1)=2(1)-\frac{1}{(1^{1/2})} =1$ this is the slope so let's use the point slope formula.

y+1=1(x-1)

y+1=x-1

y=x-2

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AP Calculus AB : Integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Integrals

Example Question #2 : Integrals

Example Question #1 : Integrals

Example Question #1 : Techniques Of Antidifferentiation

Example Question #1 : Integrals

Example Question #1 : Techniques Of Antidifferentiation

Example Question #1 : Integrals

Example Question #8 : Integrals

Example Question #1 : Integrals

Example Question #1 : Integrals

All AP Calculus AB Resources

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