A) \[1-{{x}^{2}}\]
B) \[\frac{1}{2}(1+{{x}^{2}})\]
C) \[1+{{x}^{2}}\]
D) \[\frac{1}{2}(1-{{x}^{2}})\]
Correct Answer: B
Solution :
Given, \[f(x)=\int_{-1}^{x}{|t|}\,dt\] \[=\int_{-1}^{0}{|t|\,dt}+\int_{0}^{x}{|t|\,dt}\] \[=\int_{-1}^{0}{-t}\,dt+\int_{0}^{x}{t\,dt}\] \[=-\left[ \frac{{{t}^{2}}}{2} \right]_{-1}^{0}+\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{x}\] \[=-\left[ 0-\frac{1}{2} \right]+\left[ \frac{{{x}^{2}}}{2}-0 \right]\] \[=\frac{1}{2}(1-{{x}^{2}})\]
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