KCET · Maths · Definite Integration
If \(f(x)=\int_{-1}^{x}|t| d t\), then for any \(x \geq 0, f(x)\) is equal to
- A \(1-\mathrm{x}^{2}\)
- B \(\frac{1}{2}\left(1+\mathrm{x}^{2}\right)\)
- C \(1+x^{2}\)
- D \(\frac{1}{2}\left(1-\mathrm{x}^{2}\right)\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\left(1+\mathrm{x}^{2}\right)\)
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{f}(\mathrm{x})=\int_{-1}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt}\)
\[
\begin{aligned}
&=\int_{-1}^{0}|\mathrm{t}| \mathrm{dt}+\int_{0}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt} \\
&=\int_{-1}^{0}-\mathrm{t} \mathrm{dt}+\int_{0}^{\mathrm{x}} \mathrm{t} \mathrm{dt} \\
&=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{-1}^{0}+\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{\mathrm{x}}=-\left[0-\frac{1}{2}\right]+\left[\frac{\mathrm{x}^{2}}{2}-0\right] \\
&=\frac{1}{2}\left(1+\mathrm{x}^{2}\right)
\end{aligned}
\]
\[
\begin{aligned}
&=\int_{-1}^{0}|\mathrm{t}| \mathrm{dt}+\int_{0}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt} \\
&=\int_{-1}^{0}-\mathrm{t} \mathrm{dt}+\int_{0}^{\mathrm{x}} \mathrm{t} \mathrm{dt} \\
&=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{-1}^{0}+\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{\mathrm{x}}=-\left[0-\frac{1}{2}\right]+\left[\frac{\mathrm{x}^{2}}{2}-0\right] \\
&=\frac{1}{2}\left(1+\mathrm{x}^{2}\right)
\end{aligned}
\]
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