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MHTCET 26 Apr 2025 Shift 2 Paper

Question : 105 of 150

Marks: +1, -0

\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x=

Solution:

Concept:

The absolute value function

\left|x-\frac{1}{2}\right|

changes its form at

x=\frac{1}{2}

, so the integral must be split into two parts: from

0

\frac{1}{2}

and from

\frac{1}{2}

1

Explanation:

For

0 \le x < \frac{1}{2}

, we have

\left|x-\frac{1}{2}\right| = \frac{1}{2} - x

For

\frac{1}{2} \le x \le 1

, we have

\left|x-\frac{1}{2}\right| = x - \frac{1}{2}

Therefore, split the given integral as:

\int_{0}^{1} x\left|x-\frac{1}{2}\right| dx = \int_{0}^{1/2} x\left(\frac{1}{2}-x\right) dx + \int_{1/2}^{1} x\left(x-\frac{1}{2}\right) dx.

Simplify each integrand:

= \int_{0}^{1/2} \left(\frac{x}{2}-x^2\right) dx + \int_{1/2}^{1} \left(x^2-\frac{x}{2}\right) dx.

Find the antiderivatives:

= \left[\frac{x^2}{4} - \frac{x^3}{3}\right]_{0}^{1/2} + \left[\frac{x^3}{3} - \frac{x^2}{4}\right]_{1/2}^{1}.

Evaluate the limits:

= \left(\frac{1}{16} - \frac{1}{24}\right) + \left[\left(\frac{1}{3} - \frac{1}{4}\right) - \left(\frac{1}{24} - \frac{1}{16}\right)\right].

Simplify:

= \left(\frac{1}{16} - \frac{1}{24}\right) + \left(\frac{1}{3} - \frac{1}{4} + \frac{1}{16} - \frac{1}{24}\right) = \frac{1}{48} + \frac{5}{48} = \frac{1}{8}.

Answer:

\int_{0}^{1} x\left|x-\frac{1}{2}\right| dx = \frac{1}{8}

Hence, the correct option is C, i.e.

\frac{1}{8}

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