First slide

Evaluation of definite integrals

Question

$\int_{- 2}^{3} |x^{2} - 1| dx$ is equal to

Moderate

A

3
B

$\frac{1}{3}$
C

$\frac{17}{3}$
D

$\frac{28}{3}$

Solution

$\int_{- 2}^{3} |x^{2} - 1| dx = \int_{- 2}^{- 1} |x^{2} - 1| dx + \int_{- 1}^{1} |x^{2} - 1| dx + \int_{1}^{3} |x^{2} - 1| dx$
[here, modulus function will change at the points, ,when $x^{2} - 1 = 0 i . . , at x = \pm 1$ ]
so,
$\begin{array}{l} I = \int_{- 2}^{- 1} (x^{2} - 1) dx + \int_{- 1}^{1} (1 - x^{2}) dx + \int_{1}^{3} (x^{2} - 1) dx \\ = {[\frac{x^{3}}{3} - x]}_{- 2}^{- 1} + {[x - \frac{x^{3}}{3}]}_{- 1}^{1} + {[\frac{x^{3}}{3} - x]}_{1}^{3} \\ = \frac{2}{3} + \frac{2}{3} + \frac{2}{3} + \frac{2}{3} + 6 + \frac{2}{3} = \frac{28}{3} \end{array}$

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