Q.16 - Calculus

Question

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) =$\sqrt{4-x^2}\text{ and }g\left(x\right)=-\sqrt{4-x^2}\text{ on }[-2,2]\text{. }$

Step-by-Step Solution

Answer

Part (a) 0

Part (b) 0

Part (c) 12.56

Part (a) Step 1. Given information is:

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

The objective is to find the signed area on $[-\pi ,\pi ]$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,$f\left(x\right)=\cos x$

The objective is to find the average value on $[0,2\pi ]$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:

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

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

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{graph}: \\ \mathrm{It} \mathrm{is} \mathrm{a} \mathrm{circle} \mathrm{with} \mathrm{radius}=4. \mathrm{Hence}, \mathrm{area} \mathrm{is}: \\ \mathrm{Area}=\mathrm{\pi }{\left(2\right)}^2 \mathrm{Area}=\mathrm{\pi }\left(4\right) \\ \mathrm{Area}=3.14 \times 4 \\ \mathrm{Area}=12.56 \end{gathered}$$