Q. 34 - Calculus

Question

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

$\left(a\right)$ the signed area and

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

$f\left(x\right)=2-e^{-x}, [a,b]=[-1,0]$

Step-by-Step Solution

Answer

The answer of

Part (a) The signed is $0.282$

Part (b) The absolute area is $0.282$

Part (a) Step 1. Given Information.

The given function and interval is $f\left(x\right)=2-e^{-x}, [a,b]=[-1,0]$.

Part (a) Step 2. Explanation.

The signed area in the interval will be,

$$\begin{gathered} {\int }_{-1}^0\left(2-e^{-x}\right)dx \\ ={\int }_{-1}^{0}2dx-{\int }_{-1}^0{e}^{-x}dx \\ =2[x{]}_{-1}^0+{\left[e^{-x}\right]}_{-1}^0 \\ =2+1-e \\ =3-e \\ \approx 0.282 \end{gathered}$$

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,

$$\begin{gathered} {\int }_{-1}^0\left(2-e^{-x}\right)dx \\ =-{\int }_{-1}^{-0.7}\left(2-e^{-x}\right)dx+{\int }_{-0.7}^0\left(2-e^{-x}\right)dx \\ =-\left[2[x{]}_{-1}^{-0.7}+{\left[e^{-x}\right]}_{-1}^{-0.7}\right]+\left[2[x{]}_{-0.7}^0+{\left[e^{-x}\right]}_{-0.7}^0\right] \\ \approx -0.105+0.387 \\ \approx 0.282 \end{gathered}$$