Question

${\int }_a^{b}si{n}^{2}xdx$ lies in which of the following interval if a,b $ϵ$ R and b > a ?

• A. [a, b]
• B. [0, a]
• C. [0, b ]
• D. [0, (b - a )]

Answer 1

The correct option is D [0, (b - a )]

Whenever we are asked to find the interval in which an integral is lying we have to see the interval in which the integrand is lying.
Here, we have $si{n}^{2}x$ as integrand. We know that $si{n}^{2}x$ lies in the interval [0,1].
So, $0\le si{n}^{2}x\le 1$
Also we know if a function f(x) lies between [m , M]
I.e. if $m\le f\left(x\right)\le M$
Then $(b-a)m\le {\int }_a^{b}f\left(x\right)dx\le (b-a)M$
So, using this we can say that -
$(b-a).0\le {\int }_a^{b}si{n}^2\left(x\right)dx\le (b-a).1=0\le {\int }_a^{b}si{n}^2\left(x\right)dx\le (b-a)$
So, ${\int }_a^{b}si{n}^2\left(x\right)dx$ will lie in the interval [0, (b - a )