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Q.16

Expert-verified
Found in: Page 385

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Without calculating any sums or definite integrals, determine the values of the described quantities. (Hint: Sketch graphs first.)(a) The signed area between the graph of f(x) = cos x and the x-axis on [−π, π].(b) The average value of f(x) = cos x on [0, 2π].(c) The area of the region between the graphs of f(x) =$\sqrt{4-{x}^{2}}\text{and}g\left(x\right)=-\sqrt{4-{x}^{2}}\text{on}\left[-2,2\right]\text{.}$

Part (a) 0

Part (b) 0

Part (c) 12.56

See the step by step solution

## Part (a) Step 1. Given information is:

The function is,$f\left(x\right)=\mathrm{cos}x$.

The objective is to find the signed area on $\left[-\pi ,\pi \right]$without calculating.

## Part (a) Step 2. Plotting Graph and Calculating Area

Consider the following graph of the function:

From the graph it is seen that two areas are negative while two areas are positive of equal areas.

Therefore, the signed area is 0.

## Part (b) Step 1. Given information is:

The function is,$f\left(x\right)=\mathrm{cos}x$

The objective is to find the average value on $\left[0,2\pi \right]$without calculating.

## Part (b) Step 2. Plotting Graph and Calculating area

From the above graph it is seen that both the positive area bis equal to the negative area.

Therefore, the average value is 0.

## Part (c) Step 1. Given information is:

$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{4-{\mathrm{x}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{g}\left(\mathrm{x}\right)=-\sqrt{4-{\mathrm{x}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{objective}\mathrm{is}\mathrm{to}\mathrm{find}\mathrm{area}\mathrm{between}\mathrm{f}\left(\mathrm{x}\right)\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)\mathrm{on}\left[-2,2\right]$

## Part (c) Step 2. Plotting Graph and Calculating Area

$\mathrm{Consider}\mathrm{the}\mathrm{graph}:\phantom{\rule{0ex}{0ex}}\mathrm{It}\mathrm{is}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{radius}=4.\mathrm{Hence},\mathrm{area}\mathrm{is}:\phantom{\rule{0ex}{0ex}}\mathrm{Area}=\mathrm{\pi }{\left(2\right)}^{2}\mathrm{Area}=\mathrm{\pi }\left(4\right)\phantom{\rule{0ex}{0ex}}\mathrm{Area}=3.14×4\phantom{\rule{0ex}{0ex}}\mathrm{Area}=12.56$