Q. 6 - Calculus

Question

Suppose $f\left(x\right)\ge g\left(x\right)$ on [1, 3] and $f\left(x\right)\le g\left(x\right)$ on (−∞, 1] and [3,∞). Write the area of the region between the graphs of f and g on [−2, 5] in terms of definite integrals without using absolute values .

Step-by-Step Solution

Answer

The area is${\int }_{-2}^1 \left(g\right(x)-f(x\left)\right)dx+{\int }_1^3 \left(f\right(x)-g(x\left)\right)dx+{\int }_3^5 \left(g\right(x)-f(x\left)\right)dx.$

Step 1. Given Information

It is supposed that $f\left(x\right)\ge g\left(x\right)\text{ on }[1,3]\text{ and }f\left(x\right)\le g\left(x\right)\text{ on }(-\mathrm{\infty },1]\text{ and }[3,\mathrm{\infty })\text{. }$

A function f is positive on $(-\mathrm{\infty },-1];[2,\mathrm{\infty })$ and negative on $[-1,2].$

The objective is to write the area on $[-2,5].$

The intervals will be $[-2,1],[1,3],[3,5].$

Step 2. Area,

The absolute area will be,

${\int }_{-2}^1 \left(g\right(x)-f(x\left)\right)dx+{\int }_1^3 \left(f\right(x)-g(x\left)\right)dx+{\int }_3^5 \left(g\right(x)-f(x\left)\right)dx$

Therefore, the area is ${\int }_{-2}^1 \left(g\right(x)-f(x\left)\right)dx+{\int }_1^3 \left(f\right(x)-g(x\left)\right)dx+{\int }_3^5 \left(g\right(x)-f(x\left)\right)dx.$