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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

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45 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Use Riemann Sums

The function $f\left ( x \right )$ has the following values on the interval $x=\left [ -2,13 \right ]$ :

$f\left ( -2 \right )=5$

$f\left ( 1 \right )=-19$

$f\left ( 4 \right )=5$

$f\left ( 7 \right )=-6$

$f\left ( 10 \right )=-18$

$f\left ( 13 \right )=0$

Approximate the integral of $f\left ( x \right )$ on this interval using right Riemann sums.

Possible Answers:

495

Correct answer:

Explanation:

A Riemann sum integral approximation over an interval $\left [ a, b \right ]$ with subintervals follows the form:

$\int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of rectangles, each with a base of length: $\frac{b-a}{n}$ and variable heights: $f(x_{i})$ , which depend on the function value at $x_{i}$ .

In our case, we have function values over an interval:

$f\left ( -2 \right )=5$

$f\left ( 1 \right )=-19$

$f\left ( 4 \right )=5$

$f\left ( 7 \right )=-6$

$f\left ( 10 \right )=-18$

$f\left ( 13 \right )=0$

Although the total interval has a length of .

Notice the points are equidistant. This distance is our subinterval and since we are using the right point of each interval, we disregard the first function value:

$\int_{-2}^{13}f\left ( x \right )\approx 3\left [ (-19)+(5)+(-6)+(-18)+(0)\right ]$

$\int_{-2}^{13}f\left ( x \right )\approx -114$

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Example Question #4 : Integrating

Approximate $\int_{-8}^{1}f(x)$

Using a right Riemann sum, if f(x) has the values:

f(-8)=-13

f(-5)=-19

f(-2)=6

f(1)=-9

Possible Answers:

Correct answer:

Explanation:

A Riemann sum integral approximation over an interval $\left [ a, b \right ]$ with subintervals follows the form:

$\int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of rectangles, each with a base of length: $\frac{b-a}{n}$ and variable heights: $f(x_{i})$ , which depend on the function value at $x_{i}$ .

In our case, we have function values over an interval:

f(-8)=-13

f(-5)=-19

f(-2)=6

f(1)=-9

Although the total interval has a length of notice the points are equidistant. This distance is our subinterval: . Since we are using the right point of each interval, we disregard the first function value:

$\int_{-8}^{1}f(x)\approx 3\left [ (-19)+(6)+(-9) \right ]$

$\int_{-8}^{1}f(x)\approx -66$

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Example Question #5 : Integrating

Approximate $\int_{-2}^{23}f(x)$

Using a right Riemann sum, if f(x) has the values:

f(-2)=-2

f(3)=-15

f(8)=14

f(13)=15

f(18)=-9

f(23)=-12

Possible Answers:

Correct answer:

Explanation:

A Riemann sum integral approximation over an interval $\left [ a, b \right ]$ with subintervals follows the form:

$\int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of rectangles, each with a base of length: $\frac{b-a}{n}$ and variable heights: $f(x_{i})$ , which depend on the function value at $x_{i}$ .

In our case, we have function values over an interval:

f(-2)=-2

f(3)=-15

f(8)=14

f(13)=15

f(18)=-9

f(23)=-12

Although the total interval has a length of notice the points are equidistant, with intervals between. Each equidistant interval has length: 5

Since we are using the right point of each interval, we disregard the first function value:

$\int_{-2}^{23}f(x)\approx 5\left [ (-15)+(14)+(15)+(-9)+(-12) \right ]$

$\int_{-2}^{23}f(x)\approx -35$

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Example Question #1 : Use Riemann Sums

Approximate $\int_{-7}^{13}f(x)$

Using a right Riemann sum, if f(x) has the values:

f(-7)=18

f(-2)=-17

f(3)=-16

f(8)=-15

f(13)=-14

Possible Answers:

Correct answer:

Explanation:

A Riemann sum integral approximation over an interval $\left [ a, b \right ]$ with subintervals follows the form:

$\int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of rectangles, each with a base of length: $\frac{b-a}{n}$ and variable heights: $f(x_{i})$ , which depend on the function value at $x_{i}$ .

In our case, we have function values over an interval:

f(-7)=18

f(-2)=-17

f(3)=-16

f(8)=-15

f(13)=-14

Although the total interval has a length of notice the points are equidistant, with intervals between. Each equidistant interval has length: 5

Since we are using the right point of each interval, we disregard the first function value:

$\int_{-7}^{13}f(x)\approx 5\left [ (-17)+(-16)+(-15)+(-14) \right ]$

$\int_{-7}^{13}f(x)\approx -310$

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Example Question #7 : Integrating

Approximate $\int_{4}^{12}f(x)$

Using a right Riemann sum, if f(x) has the values:

f(4)=9

f(6)=10

f(8)=-18

f(10)=15

f(12)=18

Possible Answers:

272

Correct answer:

Explanation:

A Riemann sum integral approximation over an interval $\left [ a, b \right ]$ with subintervals follows the form:

$\int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of rectangles, each with a base of length: $\frac{b-a}{n}$ and variable heights: $f(x_{i})$ , which depend on the function value at $x_{i}$ .

In our case, we have function values over an interval:

f(4)=9

f(6)=10

f(8)=-18

f(10)=15

f(12)=18

Although the total interval has a length of notice the points are equidistant, with intervals between. Each equidistant interval has length: . Since we are using the right point of each interval, we disregard the first function value:

$\int_{4}^{12}f(x)\approx 2\left [ (10)+(-18)+(15)+(18) \right ]$

$\int_{4}^{12}f(x)\approx 50$

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Example Question #8 : Integrating

The function $f\left ( x \right )$ has the following values on the interval $x=\left [ -2,38 \right ]$ :

$f\left ( -2 \right )=-13$

$f\left ( 6 \right )=-4$

$f\left ( 14 \right )=-15$

$f\left ( 22 \right )=-19$

$f\left ( 30 \right )=18$

$f\left ( 38 \right )=-8$

Approximate the integral of $f\left ( x \right )$ on this interval using right Riemann sums.

Possible Answers:

Correct answer:

Explanation:

A Riemann sum integral approximation over an interval $\left [ a, b \right ]$ with subintervals follows the form:

$\int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of rectangles, each with a base of length: $\frac{b-a}{n}$ and variable heights: $f(x_{i})$ , which depend on the function value at $x_{i}$ .

In our case, we have function values over an interval:

$f\left ( -2 \right )=-13$

$f\left ( 6 \right )=-4$

$f\left ( 14 \right )=-15$

$f\left ( 22 \right )=-19$

$f\left ( 30 \right )=18$

$f\left ( 38 \right )=-8$

Although the total interval has a length of .

Notice the points are equidistant, with intervals between them. Each equidistance interval has length: . Since we are using the right point of each interval, we disregard the first function value:

$\int_{-2}^{38}f\left ( x \right )\approx 8\left [ (-4)+(-15)+(-19)+(18)+(-8)\right ]$

$\int_{-2}^{38}f\left ( x \right )\approx -224$

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Example Question #9 : Integrating

Approximate $\int_{0}^{15}f(x)$

Using a right Riemann sum, if f(x) has the values:

f(0)=-19

f(5)=14

f(10)=2

f(15)=15

Possible Answers:

155

180

Correct answer:

155

Explanation:

A Riemann sum integral approximation over an interval $\left [ a, b \right ]$ with subintervals follows the form:

$\int_{a}^{b}f\left ( x \right )dx\approx \frac{b-a}{n}(f(x_{1}))+f\left ( x_{2} \right )+\cdots +f(x_{n}))$

It is essentially a sum of rectangles, each with a base of length: $\frac{b-a}{n}$ and variable heights: $f(x_{i})$ , which depend on the function value at $x_{i}$ .

In our case, we have function values over an interval:

f(0)=-19

f(5)=14

f(10)=2

f(15)=15

$\int_{0}^{15}f(x)\approx 5\left [ (14)+(2)+(15) \right ]$

$\int_{0}^{15}f(x)\approx 155$

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Example Question #1 : Understand The Fundamental Theorem Of Calculus

Use the fundamental theorem of Calculus to evaluate the definite integral

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

Possible Answers:

$\frac{\pi}{2}-2$

$\pi-2$

$2+\pi$

$2-\frac{\pi]}{2}$

Correct answer:

$\frac{\pi}{2}-2$

Explanation:

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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Example Question #1 : Understand The Fundamental Theorem Of Calculus

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

Possible Answers:

$\frac{-2}{(x-1)^2}$

$\frac{2(x-1)}{x}$

$\frac{-2}{x^2}$

$\frac{2}{x^2}$

$\frac{-2(x-1)}{x}$

Correct answer:

$\frac{2(x-1)}{x}$

Explanation:

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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Example Question #481 : Calculus Ab

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

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

Possible Answers:

ln6-ln3

ln2

$\frac{1}{2}$

ln3

$\frac{1}{6}-\frac{1}{3}$

Correct answer:

ln2

Explanation:

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 F(x) 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 lnx , so we must evaluate ln6-ln3 .

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 ln2.

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Use Riemann Sums

Example Question #4 : Integrating

Example Question #5 : Integrating

Example Question #1 : Use Riemann Sums

Example Question #7 : Integrating

Example Question #8 : Integrating

Example Question #9 : Integrating

Example Question #1 : Understand The Fundamental Theorem Of Calculus

Example Question #1 : Understand The Fundamental Theorem Of Calculus

Example Question #481 : Calculus Ab

All Calculus AB Resources

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