Integrals

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AP Calculus BC › Integrals

Questions 1 - 10

Find the volume of the solid generated by rotating about the y-axis the region under the curve $y =\sqrt{x}$ , from to .

$\frac{128}{5}\pi$

$\frac{64}{3}\pi$

$\frac{\pi}{2}$

$\frac{256}{7}\pi^2$

None of the other answers

Explanation

Since we are revolving a function of around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

$\int_{0}^{4}2\pi x (\sqrt{x})dx$

$=\int_{0}^{4}2\pi x^{3/2}dx$

$=[2\pi x^{5/2}]^{4}_{0}$

$=\frac{128}{5}\pi$ .

Find the volume of the solid generated by rotating about the y-axis the region under the curve $y =\sqrt{x}$ , from to .

$\frac{128}{5}\pi$

$\frac{64}{3}\pi$

$\frac{\pi}{2}$

$\frac{256}{7}\pi^2$

None of the other answers

Explanation

Since we are revolving a function of around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

$\int_{0}^{4}2\pi x (\sqrt{x})dx$

$=\int_{0}^{4}2\pi x^{3/2}dx$

$=[2\pi x^{5/2}]^{4}_{0}$

$=\frac{128}{5}\pi$ .

Find the volume of the solid generated by rotating about the y-axis the region under the curve $y =\sqrt{x}$ , from to .

$\frac{128}{5}\pi$

$\frac{64}{3}\pi$

$\frac{\pi}{2}$

$\frac{256}{7}\pi^2$

None of the other answers

Explanation

Since we are revolving a function of around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

$\int_{0}^{4}2\pi x (\sqrt{x})dx$

$=\int_{0}^{4}2\pi x^{3/2}dx$

$=[2\pi x^{5/2}]^{4}_{0}$

$=\frac{128}{5}\pi$ .

Evaluate $\int_{0}^{6} (x-3)^2 dx$

$\frac{1}{3}$

Explanation

$\int (x-3)^2 dx = \frac{1}{3} (x-3)^3 + C$

Use the fundamental theorem of calculus to evaluate:

$\frac{1}{3} (6-3)^3 - [ \frac{1}{3}(0-3)^3] = \frac{1}{3} (3)^3 - \frac{1}{3} (-3)^3$

$9 + \frac{1}{3}(27) = 9+9 = 18$

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\&\int_{0}^{12}(11sin(20sin(x)))dx\&\text{Using }3\text{ intervals.}\end{align*}$

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{0}^{12}(11sin(20sin(x)))dx\&\text{So the interval is }[0,12]\text{ the subintervals have length }\frac{12-(0)}{3}=4\&\text{and since we are using the *left*point of each interval, the x-values are:}\&[0,4,8]\&\int_{0}^{12}(11sin(20sin(x)))dx=4[(0)+(11sin(20sin(4)))+(11sin(20sin(8)))]\&\int_{0}^{12}(11sin(20sin(x)))dx=11.66\end{align*}$

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\&\int_{0}^{12}(11sin(20sin(x)))dx\&\text{Using }3\text{ intervals.}\end{align*}$

Explanation

Evaluate $\int_{0}^{6} (x-3)^2 dx$

$\frac{1}{3}$

Explanation

$\int (x-3)^2 dx = \frac{1}{3} (x-3)^3 + C$

Use the fundamental theorem of calculus to evaluate:

$\frac{1}{3} (6-3)^3 - [ \frac{1}{3}(0-3)^3] = \frac{1}{3} (3)^3 - \frac{1}{3} (-3)^3$

$9 + \frac{1}{3}(27) = 9+9 = 18$

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx\&\text{Using left points over }4\text{ intervals.}\end{align*}$

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx\&\text{So the interval is }[-2,11.6]\text{ the subintervals have length }\frac{11.6-(-2)}{4}=\frac{17}{5}\&\text{and since we are using the *left*point of each interval, the x-values are:}\&[-2,\frac{7}{5},\frac{24}{5},\frac{41}{5}]\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=\frac{17}{5}[(18\cdot 3^{(4tan(8))})+(\frac{18}{3^{(4tan(\frac{28}{5}))}})+(\frac{18}{3^{(4tan(\frac{96}{5}))}})+(\frac{18}{3^{(4tan(\frac{164}{5}))}})]\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=2200.80\end{align*}$

Evaluate $\int_{0}^{6} (x-3)^2 dx$

$\frac{1}{3}$

Explanation

$\int (x-3)^2 dx = \frac{1}{3} (x-3)^3 + C$

Use the fundamental theorem of calculus to evaluate:

$\frac{1}{3} (6-3)^3 - [ \frac{1}{3}(0-3)^3] = \frac{1}{3} (3)^3 - \frac{1}{3} (-3)^3$

$9 + \frac{1}{3}(27) = 9+9 = 18$

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx\&\text{Using left points over }4\text{ intervals.}\end{align*}$

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx\&\text{So the interval is }[-2,11.6]\text{ the subintervals have length }\frac{11.6-(-2)}{4}=\frac{17}{5}\&\text{and since we are using the *left*point of each interval, the x-values are:}\&[-2,\frac{7}{5},\frac{24}{5},\frac{41}{5}]\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=\frac{17}{5}[(18\cdot 3^{(4tan(8))})+(\frac{18}{3^{(4tan(\frac{28}{5}))}})+(\frac{18}{3^{(4tan(\frac{96}{5}))}})+(\frac{18}{3^{(4tan(\frac{164}{5}))}})]\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=2200.80\end{align*}$

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