Q. 37 - Calculus

Question

For each function $f$f and interval $[a,b]$ in Exercises 26–37, use definite integrals and the Fundamental Theorem of Calculus to find the exact values of

(a) the signed area and 

(b) the absolute area of the region between the graph of f and the x-axis from $x=a and x=b.$

$f\left(x\right)=\frac{1}{1+x^2}, [a,b]=[-1,1]$

Step-by-Step Solution

Answer

The answer is 

Part (a) The signed area is $\frac{\mathrm{\pi }}{2}$.

Part (b) The absolute area is $\frac{\mathrm{\pi }}{2}$.

Part (a) Step 1. Given Information.

The given function and interval is $f\left(x\right)=\frac{1}{1+x^2}, [a,b]=[-1,1]$.

Part (a) Step 2. Explanation .

The signed area in the interval will be,  

$$\begin{gathered} {\int }_{-1}^1\left(\frac{1}{1+x^2}\right)dx \\ ={\tan }^{-1}\left(x\right) \\ ={\tan }^{-1}\left(1\right)-{\tan }^{-1}(-1) \\ =\frac{\pi }{4}+\frac{\pi }{4} \\ =\frac{2\pi }{4} \\ =\frac{\pi }{2} \end{gathered}$$

Part (b) Step 1. Graph of the function

The graph of the function is,  

Part (b) Step 2. Absolute area.

The absolute area will be,  

$$\begin{gathered} {\int }_{-1}^1\left(\frac{1}{1+x^2}\right)dx \\ ={\tan }^{-1}\left(x\right) \\ =\frac{\pi }{4}+\frac{\pi }{4} \\ =\frac{\pi }{2} \end{gathered}$$