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AP Calculus BC : AP Calculus BC

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Example Questions

Example Question #13 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx\\&\text{Using midpoints over }3\text{ intervals.}\end{align*}$

Possible Answers:

-21.98

-12.41

-3.55

-0.40

Correct answer:

-3.55

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx\\&\text{So the interval is }[-2,2.5]\text{ the subintervals have length }\frac{2.5-(-2)}{3}=\frac{3}{2}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{5}{4},\frac{1}{4},\frac{7}{4}]\\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx=\frac{3}{2}[(-cos(11e^{(-\frac{5}{2})}))+(-cos(11e^{(\frac{1}{2})}))+(-cos(11e^{(\frac{7}{2})}))]\\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx=-3.55\end{align*}$

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Example Question #151 : Ap Calculus Bc

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

4.60

0.53

1.18

23.92

Correct answer:

4.60

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{4.8}(-tan(14sin(3x)))dx\\&\text{So the interval is }[0,4.8]\text{ the subintervals have length }\frac{4.8-(0)}{4}=\frac{6}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{3}{5},\frac{9}{5},3,\frac{21}{5}]\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx=\frac{6}{5}[(-tan(14sin(\frac{9}{5})))+(-tan(14sin(\frac{27}{5})))+(-tan(14sin(9)))+(-tan(14sin(\frac{63}{5})))]\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx=4.60\end{align*}$

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Example Question #43 : Integrals

Approximate

$\int_{1}^{2} e ^{x^{2}} \mathrm{d}x$

using the trapezoidal rule with n=5 . Round your answer to three decimal places.

Possible Answers:

10.502

20.878

14.644

31.379

15.690

Correct answer:

15.690

Explanation:

The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_{1}^{2} e ^{x^{2}} \mathrm{d}x$ by evaluating

$T = \frac{\Delta x}{2} \left [ f(x_{0}) + 2 f(x_{1}) + 2 f(x_{2}) + 2 f(x_{3}) + 2 f(x_{4}) + f(x_{5}) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^{x^{2}}$

and

$x_{0} = 1, x_{1} = 1.2,...x_{5} = 2$ .

$T =\frac{0.2}{2} \left [ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right ]$

$f(1) = e ^{1^{2}} \approx 2.7183$

$f(1.2) = e ^{1.2^{2}} \approx 4.2207$

$f(1.4) = e ^{1.4^{2}} \approx 7.0993$

$f(1.6) = e ^{1.6^{2}} \approx 12.9358$

$f(1.8) = e ^{1.8^{2}} \approx 25.5337$

$f(2) = e ^{2^{2}} \approx 54.5982$

$T \approx \frac{0.2}{2} \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

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Example Question #152 : Ap Calculus Bc

Approximate

$\int_{0}^{ \frac{\pi}{4}} \tan ^{2}x \; \mathrm{d}x$

using the trapezoidal rule with n = 4 . Round your answer to three decimal places.

Possible Answers:

0.258

0.454

0.325

0.227

0.129

Correct answer:

0.227

Explanation:

The interval $\left [ 0, \frac{\pi }{4}\right ]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = \frac{\pi }{4} \div 4 = \frac{\pi }{16}$ units in width. They are

$\left [ 0, \frac{\pi }{16} \right ], \left [ \frac{\pi }{16}, \frac{\pi }{8} \right ], \left [ \frac{\pi }{8}, \frac{3 \pi }{16} \right ], \left [ \frac{3 \pi }{16}, \frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_{0}^{ \frac{\pi}{4}} \tan ^{2}x \; \mathrm{d}x$ by evaluating

$T = \frac{\Delta x}{2} \left [ f(x_{0}) + 2 f(x_{1}) + 2 f(x_{2}) + 2 f(x_{3}) + 2 f(x_{4}) + f(x_{5}) \right ]$ ,

where

$\Delta x = \frac{\pi }{16}$ ,

$f (x) = \tan ^{2}x$ ,

and

$x_{0} = 0, x_{1} = \frac{\pi }{16},...x_{4} = \frac{\pi }{4}$ .

$T =\frac{\frac{\pi }{16}}{2} \left [ f(0) + 2 f \left ( \frac{ \pi }{16}\right ) + 2 f \left ( \frac{ \pi }{8}\right ) + 2 f \left ( \frac{ 3 \pi }{16}\right ) + f \left ( \frac{ \pi }{4} \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( \frac{ \pi }{16}\right ) = \tan ^{2} \frac{ \pi }{16} \approx 0.0396$

$f\left ( \frac{ \pi }{8}\right ) = \tan ^{2} \frac{ \pi }{8} \approx 0.1716$

$f\left ( \frac{ 3 \pi }{16}\right ) = \tan ^{2} \frac{ 3 \pi }{16} \approx 0.4465$

$f\left ( \frac{ \pi }{4}\right ) = \tan ^{2} \frac{ \pi }{4} = 1$

$\frac{\frac{\pi }{16}}{2} = \frac{\pi }{32} = 0.0982$

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

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Example Question #44 : Integrals

Approximate

$\int_{1}^{5} \frac{ \ln x \; \mathrm {d}x }{x^2}$

using the trapezoidal rule with n = 4 . Round your estimate to three decimal places.

Possible Answers:

0.382

0.828

0.414

0.503

0.446

Correct answer:

0.414

Explanation:

The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width - they are [1,2], [2,3],[3,4], [4,5] .

The trapezoidal rule approximates the area of the given integral $\int_{1}^{5} \frac{ \ln x \; \mathrm {d}x }{x^2}$ by evaluating

$T = \frac{\Delta x}{2} \left [ f(x_{0}) + 2 f(x_{1}) + 2 f(x_{2}) + 2 f(x_{3}) + f(x_{4}) \right ]$ ,

where $\Delta x = 1$ , $f(x) = \frac{ \ln x }{x^2}$ , and

$x_{0} = 1, x_{1} = 2, ... x_{4} = 5$ .

$T =\frac{1}{2} \left [ f(1) + 2 f(2}) + 2 f(3) + 2 f(4) + f(5) \right ]$

$f(1) = \frac{ \ln 1 }{1^2} = 0$

$f(2) = \frac{ \ln 2 }{2^2} \approx \frac{0.6931 }{4} \approx 0.1733$

$f(3) = \frac{ \ln 3 }{3^2} \approx \frac{1.0986 }{9} \approx 0.1221$

$f(4) = \frac{ \ln 4 }{4^2} \approx \frac{1.3863 }{16} \approx 0.0866$

$f(5) = \frac{ \ln 5 }{5^2} \approx \frac{1.6094 }{25} \approx 0.0644$

$T \approx 0.5 \left [0+ 2 \cdot 0.1733 + 2 \cdot 0.1221 + 2 \cdot 0.0866 + 0.0644 \right ]$

$\approx 0.5 \left [ 0.3466 + 0.2442 + 0.1732 + 0.0644 \right ]$

$\approx 0.5 \cdot 0.8284 \approx 0.414$

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Example Question #1 : Fundamental Theorem Of Calculus And Techniques Of Antidifferentiation

Find the result:

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{e^x} t \ln t \; \mathrm{d}t$

Possible Answers:

$xe^{x }+ e$

2xe^x

$xe^{x+1}$

2xe^x + 1

$xe^{2x}$

Correct answer:

$xe^{2x}$

Explanation:

Set u = e^x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = e^x$ , and by the chain rule,

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{e^x} t \ln t \; \mathrm{d}t$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{1}^{u} t \ln t \; \mathrm{d}t$

$= e^x \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{1}^{u} t \ln t \; \mathrm{d}t$

By the fundamental theorem of Calculus, the above can be rewritten as

$e^x \cdot u \ln u$

$= e^x \cdot e^x \ln e^x$

$= e^{2x} \cdot x = xe^{2x}$

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Example Question #2 : Fundamental Theorem Of Calculus And Techniques Of Antidifferentiation

Evaluate f'(0) :

$f(x)=\int_{0}^{x} \sin\left(\sqrt{t^2+\frac{\pi^2}{4}} \right )dt$

Possible Answers:

$\frac{\pi}{2}$

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

Correct answer:

Explanation:

By the Fundamental Theorem of Calculus, we have that ${f'(x) = \frac{d}{dx} \int_0^x \sin \sqrt{t^2 + \frac{\pi ^2}{4}} dt = \sin \sqrt{x^2 + \frac{\pi ^2}{4}}$ . Thus, $f'(0) = \sin \sqrt{0^2 + \frac{\pi ^2}{4}} =1$ .

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Example Question #1 : Fundamental Theorem Of Calculus

Evaluate f'(4) when $f(x)=\int_{0}^{x}(2t^{2}+5t-4)dt$ .

Possible Answers:

Correct answer:

Explanation:

Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_{0}^{x}(2t^{2}+5t-4)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(2t^{2}+5t-4)dt=2x^{2}+5x-4$ .

Therefore $f'(4)=2(4)^{2}+5(4)-4=32+20-4=52-4=48$ .

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Example Question #2 : Fundamental Theorem Of Calculus

Evaluate f'(1) when $f(x)=\int_{0}^{x}(6t^{2}-3t+10)dt$ .

Possible Answers:

Correct answer:

Explanation:

Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_{0}^{x}(6t^{2}-3t+10)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(6t^{2}-3t+10)dt=6x^{2}-3x+10$ . Therefore, $f'(1)=6(1)^{2}-3(1)+10=6-3+10=3+10=13$ .

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Example Question #2 : Fundamental Theorem Of Calculus

Suppose we have the function

$\small g(x)=\int_{0}^{\cos x}(1+\sin t)dt$

What is the derivative, $\small g'(x)$ ?

Possible Answers:

$\small \small -\sin x\left[1+\sin\left(\cos(x) \right )\right]$

$\small \small \small \small \sin x\left[1+\sin\left(\cos x \right )\right]$

$\small \small \small 1+\sin\left(\cos(x) \right )$

$\small \small \small 1-\sin\left(\cos(x) \right )$

Correct answer:

$\small \small -\sin x\left[1+\sin\left(\cos(x) \right )\right]$

Explanation:

We can view the function $\small g(x)$ as a function of $\small \cos x$ , as so

$\small g(x)=F(\cos x)$

where $\small F(x)=\int_0^x (1+\sin t)dt$ .

We can find the derivative of $\small g(x)$ using the chain rule:

$\small g'(x)=F'(\cos x)\cdot (\cos x)'=F'(\cos x)\cdot (-\sin x)$

where $\small F'(z)$ can be found using the fundamental theorem of calculus:

$\small \small F'(x)=\frac{d}{dz}\int_0^z (1+\sin t)dt=1+\sin z$

So we get

$\small \small \small \small g'(x)=-\sin x\cdot F'(\cos x)=-\sin x\cdot[1+\sin(\cos x)]$

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #13 : Riemann Sum: Midpoint Evaluation

Example Question #151 : Ap Calculus Bc

Example Question #43 : Integrals

Example Question #152 : Ap Calculus Bc

Example Question #44 : Integrals

Example Question #1 : Fundamental Theorem Of Calculus And Techniques Of Antidifferentiation

Example Question #2 : Fundamental Theorem Of Calculus And Techniques Of Antidifferentiation

Example Question #1 : Fundamental Theorem Of Calculus

Example Question #2 : Fundamental Theorem Of Calculus

Example Question #2 : Fundamental Theorem Of Calculus

All AP Calculus BC Resources

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