Question

Suppose $y=f\left(x\right)$ and $y=g\left(x\right)$ are two functions whose graphs intersect at three points $(0,4),(2,2)$ and $(4,0)$ with $f\left(x\right)>g\left(x\right)$ for $0<x<2$ and $f\left(x\right)<g\left(x\right)$ for $2<x<4$. if ${\int }_0^4\left(f\right(x)-g(x\left)\right)dx=10$ and ${\int }_2^4\left(g\right(x)-f(x\left)\right)dx=5$, the area between two curves for $0<x<2$, is:

• A. 5
• B. 10
• C. 15
• D. 20

Answer 1

Answer: (C)

15

$f\left(x\right)>g\left(x\right)$ for $0<x<2$ and $f\left(x\right)<g\left(x\right)$ for $2<x<4$.
if ${\int }_0^4\left[f\right(x)-g(x\left)\right]dx=10$ and ${\int }_2^4\left[g\right(x)-f(x\left)\right)]dx=5$,
${\int }_0^4\left[f\right(x)-g(x\left)\right]dx=10$
${\int }_0^{2}f\left(x\right)-g\left(x\right)dx+{\int }_2^{4}f\left(x\right)-g\left(x\right)dx=10$
${\int }_0^2\left(f\right(x)-g(x\left)\right)dx+{\int }_2^4\left(g\right(x)-f(x\left)\right)dx=10$
${\int }_0^2\left(f\right(x)-g(x\left)\right)dx=15$