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Calculus AB : 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 : Integrating

Let $f(x) = \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x$ .

A relative maximum of the graph of can be located at:

Possible Answers:

$x =- \sqrt {6}$

$x=- \sqrt[4]{5}$

$x = \sqrt {6}$

$x= \sqrt[4]{5}$

The graph of has no relative maximum.

Correct answer:

$x =- \sqrt {6}$

Explanation:

At a relative minimum (c,f(c)) of the graph f(x) , it will hold that f'(c) = 0 and f''(c) < 0 .

First, find f'(x) . Using the sum rule,

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x \right )$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{6}{5} x^{5} \right )+ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{5}{3}x^{3} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( 30x \right )$

Differentiate the individual terms using the Constant Multiple and Power Rules:

$f'(x) = \frac{1}{7} \cdot 7 x^{6} - \frac{6}{5} \cdot 5 x^{4} + \frac{5}{3} \cdot 3 x^{2} - 30x$

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

Set this equal to 0:

$x^{6} - 6 x^{4} +5 x^{2} - 30 = 0$

$(x^{6} - 6 x^{4} )+(5 x^{2} - 30) = 0$

$x^{4} (x^{2} - 6 )+ 5 (x^{2} - 6 ) = 0$

$(x^{4}+ 5 ) (x^{2} - 6 ) = 0$

Either:

$x^{4}+5 = 0$ , in which case, $x^{4} = -5$ ; this equation has no real solutions.

$x^{2}- 6=0$ has two real solutions, $-\sqrt {6}$ and $\sqrt {6}$ .

Now take the second derivative, again using the sum rule:

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} - 6 x^{4} +5 x^{2} - 30)$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} )- \frac{\mathrm{d} }{\mathrm{d} x} ( 6 x^{4} )+\frac{\mathrm{d} }{\mathrm{d} x}( 5 x^{2} )- \frac{\mathrm{d} }{\mathrm{d} x}( 30)$

Differentiate the individual terms using the Constant Multiple and Power Rules:

$f''(x) =6 x^{5} - 6 ( 4 x^{3} )+ 5( 2x )- 0$

$f''(x) =6 x^{5} -2 4 x^{3} +10x$

Substitute $\sqrt {6}$ for :

$f''( \sqrt {6}) =6 ( \sqrt {6} )^{5} -2 4 ( \sqrt {6})^{3} +10( \sqrt {6})$

$f''(-\sqrt {6}) = 216\sqrt {6}-144 \sqrt {6} +10\sqrt {6}$

$f''(-\sqrt {6}) =82\sqrt {6} >0$

Therefore, has a relative minimum at $x = \sqrt {6}$ .

Now. substitute $-\sqrt {6}$ for :

$f''(-\sqrt {6}) =6 (-\sqrt {6} )^{5} -2 4 (-\sqrt {6})^{3} +10(-\sqrt {6})$

$f''(-\sqrt {6}) =- 216\sqrt {6}+144 \sqrt {6} -10\sqrt {6}$

$f''(-\sqrt {6}) =-82\sqrt {6} < 0$

Therefore, has a relative maximum at $x =- \sqrt {6}$ .

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Example Question #1 : Integrating

Estimate the integral of f(x)=3x^2+1 from 0 to 3 using left-Riemann sum and 6 rectangles. Use $\Delta x=0.5$

Possible Answers:

74.25

46.25

23.625

37.125

23.125

Correct answer:

23.625

Explanation:

Because our $\Delta x$ is constant, the left Riemann sum will be

$R_{L}=\Delta x(f(0)+f(0.5)+f(1)+f(1.5)+f(2)+f(2.5))$

$R_{L}=0.5\cdot (1+1.75+4+7.75+13+19.75)=23.625$

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Example Question #1 : Use Riemann Sums

Use Left Riemann sums with 4 subintervals to approximate the area between the x-axis, f(x)=x^4 +1 , x=-2 , and x=2 .

Possible Answers:

Correct answer:

Explanation:

To use left Riemann sums, we need to use the following formula:

$Area \approx \sum_{i=1}^n f(x_i)\Delta x$ .

where is the number of subintervals, (4 in our problem),

is the "counter" that denotes which subinterval we are working with,(4 subintervals mean that will be 1, 2, 3, and then 4)

f(x_i) is the function value when you plug in the "i-th" x value, (i-th in this case will be 1-st, 2-nd, 3-rd, and 4-th)

, $\Delta x$ is the width of each subinterval, which we will determine shortly.

and $\sum_{i=1}^n$ means add all versions together (for us that means add up 4 versions).

This fancy equation approximates using boxes. We can rewrite this fancy equation by writing $f(x_i)\Delta x$ , 4 times; 1 time each for i=1 , i=2 , i=3 , and i=4 . This gives us

$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

Think of $\Delta x$ as the base of each box, and f(x_1) as the height of the 1st box.

This is basically Area = (base)(height) , 4 times, and then added together.

Now we need to determine what $\Delta x, x_1, x_2, x_3,$ and x_4 are.

To find $\Delta x$ we find the total length between the beginning and ending x values, which are given in the problem as x=-2 and x=2 . We then split this total length into 4 pieces, since we are told to use 4 subintervals.

In short, $\Delta x = \frac{b - a}{n}= \frac{(2)- (-2)}{4} = 1$

Now we need to find the x values that are the left endpoints of each of the 4 subintervals. Left endpoints because we are doing Left Riemann sums.

The left most x value happens to be the smaller of the overall endpoints given in the question. In other words, since we only care about the area from x=-2 to x=2 , we'll just use the smaller one, , for our first x_i .

Now we know that x_1 = -2 .

To find the next endpoint, x_2 , just increase the first x by the length of the subinterval, which is $\Delta x = 1$ . In other words

$x_2 = x_1 + \Delta x$

x_2 = -2 + 1

x_2 = -1

Add the $\Delta x$ again to get x_3

$x_3 = x_2 + \Delta x$

x_3 = -1 + 1

x_3 = 0

And repeat to find x_4

$x_4 = x_3 + \Delta x$

x_4 = 0 + 1

x_4 = 1

Now that we have all the pieces, we can plug them in.

$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

$f(-2)\cdot1 + f(-1) \cdot (1)+ f(0)\cdot (1) + f(1)\cdot (1)$

plug each value into f(x)=x^4 +1 and then simplify.

[(-2)^4+1](1)+[(-1)^4+1](1)+ [(0)^4+1](1)+[(1)^4+1](1)

[16 + 1]+ [1 + 1]+ [0+ 1]+ [1 + 1]

17 + 2 + 1 + 2

This is the final answer, which is an approximation of the area under the function.

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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 : Integrating

Example Question #1 : Integrating

Example Question #1 : Use Riemann Sums

All Calculus AB Resources

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