Integrals - AP Calculus BC

Card 0 of 942

Question

Approximate

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

using the midpoint rule with . Round your answer to three decimal places.

Answer

The interval $\left [ 0, \frac{\pi}{2}\right ]$ is $\frac{\pi}{2}$ units in width; the interval is divided evenly into five subintervals $\Delta x = \frac{\pi}{2} \div 5 = \frac{\pi}{10}$ units in width, with their midpoints shown:

$\left [ 0, \frac{\pi}{10}\right ]; x_{1} = \frac{\pi}{20}$

$\left [ \frac{\pi}{10},\frac{\pi}{5} \right ]; x_{2} = \frac{3 \pi}{20}$

$\left [ \frac{\pi}{5},\frac{3\pi}{10} \right ]; x_{3} =\frac{5 \pi}{20} = \frac{ \pi}{4}$

$\left [ \frac{3\pi}{10}, \frac{2 \pi}{5} \right ]; x_{4} =\frac{7 \pi}{20}$

$\left [ \frac{2 \pi}{5} , \frac{\pi}{2} \right ];x_{5} = \frac{9 \pi}{20}$

The midpoint rule requires us to calculate:

$M = \Delta x \left ( f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) \right )$

where $\Delta x = \frac{\pi}{10}$ and $f(x) = x^{2} \cos x$

Evaluate $f(x) = x^{2} \cos x$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$ :

$f\left \( \frac{\pi}{20} \right\) = \left \( \frac{\pi}{20} \right\)^{2} \cos \left \( \frac{\pi}{20} \right\)$

$\approx 0.1571^{2} \cdot \cos 0.1571 \approx 0.1571^{2} \cdot 0.9877 \approx 0.0244$

$f\left \( \frac{3\pi}{20} \right\) = \left \( \frac{3\pi}{20} \right\)^{2} \cos \left \( \frac{3\pi}{20} \right\)$

$\approx 0.4712 ^{2} \cdot \cos 0.4712 \approx 0.1571^{2} \cdot 0.8910 \approx 0.1979$

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

$\approx 0.7854^{2} \cdot \cos 0.7854 \approx 0.7854^{2} \cdot 0.7071 \approx 0.4362$

$f\left \( \frac{7\pi}{20} \right\) = \left \( \frac{7\pi}{20} \right\)^{2} \cos \left \( \frac{7\pi}{20} \right\)$

$\approx 1.0996^{2} \cdot \cos 1.0996 \approx 1.0996^{2} \cdot 0.4540 \approx 0.5489$

$f\left \( \frac{9\pi}{20} \right\) = \left \( \frac{9\pi}{20} \right\)^{2} \cos \left \( \frac{9\pi}{20} \right\)$

$\approx 1.4137 ^{2} \cdot \cos1.4137 \approx 1.4137^{2} \cdot 0.1564 \approx 0.3126$

Since $\Delta x = \frac{\pi}{10} \approx 0.3142$ ,

we can approximate as

$M \approx 0.3142 \cdot \left ( 0.0244 + 0.1979 + 0.4362 + 0.5489 + 0.3126) \right ) \approx 0.478$ .

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Question

Approximate

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

using the midpoint rule with . Round your answer to three decimal places.

Answer

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, with their midpoints shown:

$\left [ 1, 1.2 \right ]; x_{1} = 1.1$

$\left [ 1.2, 1.4 \right ]; x_{2} = 1.3$

$\left [1.4, 1.6 \right ]; x_{3} =1.5$

$\left [ 1.6, 1.8 \right ]; x_{4} = 1.7$

$\left [1.8, 2 \right ];x_{5} = 1.9$

The midpoint rule requires us to calculate:

$M = \Delta x \left ( f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) \right )$

where $\Delta x = 0.2$ and $f(x) =x^{2} \ln x$

Evaluate $f(x) =x^{2} \ln x$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$ :

$f(1.1) = 1.1 ^{2} \cdot \ln 1.1 \approx 1.1 ^{2} \cdot 0.0953 \approx 0.1153$

$f(1.3) = 1.3^{2} \cdot \ln 1.3 \approx 1.3 ^{2} \cdot 0.2624 \approx 0.4434$

$f(1.5) = 1.5 ^{2} \cdot \ln 1.5 \approx 1.5 ^{2} \cdot 0.4055 \approx 0.9123$

$f(1.7) = 1.7 ^{2} \cdot \ln 1.7\approx 1.7 ^{2} \cdot 0.5306 \approx 1.5335$

$f(1.9) = 1.9 ^{2} \cdot \ln 1.9 \approx 1.9 ^{2} \cdot 0.6419 \approx 2.3171$

$M \approx 0.2 \cdot \left ( 0.1153 +0.4434 + 0.9123 + 1.5335 + 2.3171 \right ) \approx 1.064$

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Question

Approximate

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

using the midpoint rule with . Round your answer to three decimal places.

Answer

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, with their midpoints shown:

$\left [ 1,2 \right ] : x _{1} = 1.5$

$\left [2,3 \right ]: x _{2} = 2.5$

$\left [3,4 \right ] : x _{3} = 3.5$

$\left [ 4,5 \right ] : x _{4} = 4.5$

The midpoint rule requires us to calculate:

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

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

Evaluate $f(x) = \frac{ \ln x }{x^2}$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }$ :

$f(1.5) = \frac{ \ln 1.5 }{1.5^2} \approx \frac{0.4055}{2.25} \approx 0.1802$

$f(2.5) = \frac{ \ln 2.5 }{2.5^2} \approx \frac{0.9163}{6.25} \approx 0.1466$

$f(3.5) = \frac{ \ln 3.5 }{3.5^2} \approx \frac{1.2528}{12.25} \approx 0.1023$

$f(4.5) = \frac{ \ln 4.5 }{4.5^2} \approx \frac{1.5041}{20.25} \approx 0.0743$

So

$M =1 \left (0.1802 + 0.1466 +0.1023 + 0.0743 \right )$

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Question

$\begin{align}&\text{Calculate the Riemann sums integral approximation of :}\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx\&\text{Using midpoints over }4\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-4}^{14.4}(-20sin(13tan(2x)))dx\&\text{So the interval is }[-4,14.4]\text{ the subintervals have length }\frac{14.4-(-4)}{4}=\frac{23}{5}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[-\frac{17}{10},\frac{29}{10},\frac{15}{2},\frac{121}{10}]\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx=\frac{23}{5}[(20sin(13tan(\frac{17}{5})))+(-20sin(13tan(\frac{29}{5})))+(-20sin(13tan(15)))+(-20sin(13tan(\frac{121}{5})))]\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx=-159.88\end{align}$

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Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{5}^{21}(-16cos(sin(6x)))dx\&\text{Using }4\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{5}^{21}(-16cos(sin(6x)))dx\&\text{So the interval is }[5,21]\text{ the subintervals have length }\frac{21-(5)}{4}=4\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[7,11,15,19]\&\int_{5}^{21}(-16cos(sin(6x)))dx=4[(-16cos(sin(42)))+(-16cos(sin(66)))+(-16cos(sin(90)))+(-16cos(sin(114)))]\&\int_{5}^{21}(-16cos(sin(6x)))dx=-188.28\end{align}$

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Question

$\begin{align}&\text{Using }3\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{2}^{14.6}(16cos(20e^{(3x)}))dx\&\text{So the interval is }[2,14.6]\text{ the subintervals have length }\frac{14.6-(2)}{3}=\frac{21}{5}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[\frac{41}{10},\frac{83}{10},\frac{25}{2}]\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx=\frac{21}{5}[(16cos(20e^{(\frac{123}{10})}))+(16cos(20e^{(\frac{249}{10})}))+(16cos(20e^{(\frac{75}{2})}))]\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx=26.04\end{align}$

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Question

$\begin{align}&\text{Using }3\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx\&\text{So the interval is }[-3,9.9]\text{ the subintervals have length }\frac{9.9-(-3)}{3}=\frac{43}{10}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[-\frac{17}{20},\frac{69}{20},\frac{31}{4}]\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx=\frac{43}{10}[(\frac{13}{3^{(5cos(\frac{51}{20}))}})+(\frac{13}{3^{(5cos(\frac{207}{20}))}})+(\frac{13}{3^{(5cos(\frac{93}{4}))}})]\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx=7165.44\end{align}$

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Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{1}^{9}(19cos(2x^{2}))dx\&\text{Using }4\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{1}^{9}(19cos(2x^{2}))dx\&\text{So the interval is }[1,9]\text{ the subintervals have length }\frac{9-(1)}{4}=2\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[2,4,6,8]\&\int_{1}^{9}(19cos(2x^{2}))dx=2[(19cos(8))+(19cos(32))+(19cos(72))+(19cos(128))]\&\int_{1}^{9}(19cos(2x^{2}))dx=-36.91\end{align}$

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Question

$\begin{align}&\text{Using }4\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx\&\text{So the interval is }[1,12.6]\text{ the subintervals have length }\frac{12.6-(1)}{4}=\frac{29}{10}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[\frac{49}{20},\frac{107}{20},\frac{33}{4},\frac{223}{20}]\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx=\frac{29}{10}[(-9sin(18e^{(\frac{147}{20})}))+(-9sin(18e^{(\frac{321}{20})}))+(-9sin(18e^{(\frac{99}{4})}))+(-9sin(18e^{(\frac{669}{20})}))]\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx=-73.83\end{align}$

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Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{0}^{10}(8cos(19e^{(x)}))dx\&\text{Using }4\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{0}^{10}(8cos(19e^{(x)}))dx\&\text{So the interval is }[0,10]\text{ the subintervals have length }\frac{10-(0)}{4}=\frac{5}{2}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[\frac{5}{4},\frac{15}{4},\frac{25}{4},\frac{35}{4}]\&\int_{0}^{10}(8cos(19e^{(x)}))dx=\frac{5}{2}[(8cos(19e^{(\frac{5}{4})}))+(8cos(19e^{(\frac{15}{4})}))+(8cos(19e^{(\frac{25}{4})}))+(8cos(19e^{(\frac{35}{4})}))]\&\int_{0}^{10}(8cos(19e^{(x)}))dx=-44.98\end{align}$

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Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{4}^{9.4}(-15sin(x^{2}))dx\&\text{Using }3\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{4}^{9.4}(-15sin(x^{2}))dx\&\text{So the interval is }[4,9.4]\text{ the subintervals have length }\frac{9.4-(4)}{3}=\frac{9}{5}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[\frac{49}{10},\frac{67}{10},\frac{17}{2}]\&\int_{4}^{9.4}(-15sin(x^{2}))dx=\frac{9}{5}[(-15sin(\frac{2401}{100}))+(-15sin(\frac{4489}{100}))+(-15sin(\frac{289}{4}))]\&\int_{4}^{9.4}(-15sin(x^{2}))dx=2.88\end{align}$

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Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx\&\text{Using }3\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx\&\text{So the interval is }[0,12.3]\text{ the subintervals have length }\frac{12.3-(0)}{3}=\frac{41}{10}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[\frac{41}{20},\frac{123}{20},\frac{41}{4}]\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx=\frac{41}{10}[(-12sin(19e^{(\frac{41}{5})}))+(-12sin(19e^{(\frac{123}{5})}))+(-12sin(19e^{(41)}))]\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx=-70.62\end{align}$

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Question

$\begin{align}&\text{Using the method of midpoint Riemann sums approximate the integral:}\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx\&\text{Using }3\text{ intervals.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-2}^{4.6}(9tan(9tan(6x)))dx\&\text{So the interval is }[-2,4.6]\text{ the subintervals have length }\frac{4.6-(-2)}{3}=\frac{11}{5}\&\text{and since we are using the midpoint of each interval, the x-values are:}\&[-\frac{9}{10},\frac{13}{10},\frac{7}{2}]\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx=\frac{11}{5}[(-9tan(9tan(\frac{27}{5})))+(9tan(9tan(\frac{39}{5})))+(9tan(9tan(21)))]\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx=477.77\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

The temperature of an oven is increasing at a rate $r(t)=2e^{0.5t}$ degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is degrees Fahrenheit.

What is the temperture of the oven at ? Round your answer to the nearest tenth.

Answer

Integrating over an interval will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_{0}^{5}2e^{0.5t}dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When , and when Making these substitutions leads to the integral

$\int_{0}^{2.5}4e^udu$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int e^xdx=e^x+c$ .

Therefore,

$\int_{0}^{2.5}4e^udu=4e^u|_{0}^{2.5}=4e^{2.5}-4\approx 44.7$ .

This tells us that the temperature rose by approximately degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at minutes is

degrees.

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Question

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

and the velocity at time of the particle is .

Answer

In order to find the velocity function, we must integrate the accleration function:

$\int (16t^2+48)dt=\frac{16}{3}t^3+48t+C$

We used the rule

$\int x^ndx=\frac{1}{n+1}x^{n+1}+C$

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

so we plug in zero into the velocity function and solve for C:

$v(0)=30=\frac{16}{3}(0)^3+48(0)+C$

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

$v(t)=\frac{16}{3}t^3+48t+30$

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Question

Find the work done by gravity exerting an acceleration of $-10\frac{m}{s^2}$ for a block down from its original position with no initial velocity.

Remember that

$F_{grav}=mass*acceleration$

$W=\int_{a}^{b}F(x)dx$ , where is a force measured in , is work measured in , and and are initial and final positions respectively.

Answer

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= (10-10)\frac{kgm}{s^2}= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\ W=\int_{0}^{-5}-100dx\ \=\frac{-100x}{2}|_{0}^{-5} \ \=-50x|_{0}^{-5}\ \=-50(-5)--50(0)\ \=250J$

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Question

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

$f(x)=\int3x^2-e^x+2dx$

Answer

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

$f(x)=\int3x^2-e^x+2dx$

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

$1) \int x^ndx=\frac{x^{n+1}}{n+1}+c$

$2) \int e^xdx=e^x+c$

Using these two, we can integrate f(x)

$f(x)=\int3x^2-e^x+2dx$

$\int3x^2-e^x+2dx=\frac{3x^3}{3}-e^x+2x+c=x^3-e^x+2x+c$

SO, we get:

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

So, our answer is:

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Question

Evaluate:

$\int_{0}^{\infty } \frac{\cos x; \mathrm{d}x}{e^{x}}$

Answer

First, we will find the indefinite integral $\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \int e^{-x} \cos x; \mathrm{d}x$ using integration by parts.

We will let $u = \cos x$ and $\mathrm{d}v= e^{-x}; \mathrm{d}x$ .

Then $\frac{\mathrm{d} u}{\mathrm{d} x} = - \sin x$ and $\mathrm{d} u = - \sin x ; \mathrm{d} x$ .

$v= \int e^{-x}; \mathrm{d}x = - e^{-x} = - \frac{1}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \int e^{-x} \cos x; \mathrm{d}x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \cos x \cdot \left (- \frac{1}{e^{x} } \right ) - \int \left (- \frac{1}{e^{x} } \right ) ; \cdot \left (- \sin x \right ); \mathrm{d} x$

$= - \frac{\cos x}{e^{x} } - \int e^{-x} \sin x ; \mathrm{d} x$

To find $\int e^{-x} \sin x ; \mathrm{d} x$ , we use another integration by parts:

$u = \sin x$ , which means that $\mathrm{d} u = \cos x ; \mathrm{d} x$ , and

$\mathrm{d}v= e^{-x}; \mathrm{d}x$ , which means that, again, $v= - \frac{1}{e^{x} }$ .

$\int e^{-x} \sin x ; \mathrm{d} x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \sin x \cdot \left (- \frac{1}{e^{x} } \right ) - \int \left (- \frac{1}{e^{x} } \right ) ; \cdot \cos x ; \mathrm{d} x$

$= - \frac{ \sin x }{e^{x} } + \int \frac{\cos x ; \mathrm{d} x}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } - \int e^{-x} \sin x ; \mathrm{d} x$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } - \left ( - \frac{ \sin x }{e^{x} } + \int \frac{\cos x ; \mathrm{d} x}{e^{x} } \right )$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } + \frac{ \sin x }{e^{x} } - \int \frac{\cos x ; \mathrm{d} x}{e^{x} }$

$2 \int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } + \frac{ \sin x }{e^{x} }$

$2 \int \frac{\cos x; \mathrm{d}x}{e^{x}} = \frac{\sin x -\cos x}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \frac{\sin x -\cos x}{2e^{x} }$

Since

$\frac{-2}{2e^{x} } \leq \frac{\sin x -\cos x}{2e^{x} } \leq \frac{2}{2e^{x} }$ , or,

$- \frac{1}{e^{x} } \leq \frac{\sin x -\cos x}{2e^{x} } \leq \frac{1}{e^{x} }$

for all real , and

$- \lim_{x\rightarrow \infty } \frac{1}{e^{x} } = \lim_{x\rightarrow \infty } \frac{1}{e^{x} } = 0$ ,

by the Squeeze Theorem,

$\lim_{x\rightarrow \infty } \frac{\sin x -\cos x}{2e^{x} } = 0$ .

$\int_{0}^{\infty } \frac{\cos x; \mathrm{d}x}{e^{x}}$

$= \lim_{b\rightarrow \infty }\int_{0}^{b } \frac{\cos x; \mathrm{d}x}{e^{x}}$

$= \lim_{ b\rightarrow \infty }$ $\left.\begin{matrix} \frac{\sin x -\cos x}{2e^{x} } \ \textrm{ } \end{matrix}\right|$ $\begin{matrix} b\ \ 0 \end{matrix}$

$= 0 - \frac{\sin 0 -\cos 0}{2e^{0} }$

$= - \frac{ 0 -1}{2 \cdot 1 } = \frac{1}{2}$

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