Mathematics - Definite Integration Question with Solution | TestHub

If $0\le x\le \frac{1}{2}\Rightarrow f(x)={\int }_1^{x}tdt=\frac{x^2-1}{2}$ If $-\frac{1}{2}\le x\le 0\Rightarrow f(x)={\int }_1^{0}tdt-{\int }_0^{x}tdt=-\left(\frac{x^2+1}{2}\right)$

$$\begin{array}{rlr}& \Rightarrow f^1(x)>0\forall x\in \left[\frac{-1}{2},\frac{1}{2}\right]& \therefore \text{ Range of }f(x)\text{ is }\left[f\left(\frac{-1}{2}\right),f\left(\frac{1}{2}\right)\right]\Rightarrow \left[\frac{-5}{8},\frac{-3}{8}\right]\end{array}$$