Question

For a function $f\left(x\right)\ge 0\forall xϵ(a,b);a<b\left|{\int }_a^{b}f\right(x\left)dx\right|=5thenfind{\int }_a^b|f\left(x\right)|dx$ ?

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is C 5

We have seen that $\left|{\int }_a^{b}f\right(x\left)dx\right|\le {\int }_a^b|f\left(x\right)|dx$ But we are given that $f\left(x\right)\ge 0\forall xϵ(a,b)$
So, | f(x)| = f(x) . And ${\int }_a^{b}f\left(x\right)dx$ will be a positive value as f(x) is always positive in the given interval and also a < b (given). So basically both the modes are redundant.
$.{\int }_a^b|f\left(x\right)|dx={\int }_a^{b}f\left(x\right)dx$
$\&|{\int }_a^{b}f\left(x\right)dx|={\int }_a^{b}f\left(x\right)dx$
Thus, ${\int }_a^b|f\left(x\right)|dx=|{\int }_a^{b}f\left(x\right)dx|=5$
Note - If there was no information given about f(x) being positive in the interval and also about the limits
(a & b) then the answer could be 5 & 6 both. As $|{\int }_a^{b}f\left(x\right)dx|\le {\int }_a^b|f\left(x\right)|dx$ will always hold good.