मराठी

3 ∫ 1 ∣ ∣ X 2 − 2 X ∣ ∣ D X

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प्रश्न

\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]

बेरीज
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उत्तर

We have,

\[\left| x^2 - 2x \right| = \begin{cases}- \left( x^2 - 2x \right),& 1 \leq x \leq 2\\ x^2 - 2x,& 2 \leq x \leq 3\end{cases}\]

\[ \therefore \int_1^3 \left| x^2 - 2x \right| d x\]
\[ = \int_1^2 - \left( x^2 - 2x \right) dx + \int_2^3 \left( x^2 - 2x \right) dx\]

\[ = \left[ - \frac{x^3}{3} + x^2 \right]_1^2 + \left[ \frac{x^3}{3} - x^2 \right]_2^3 \]

\[ = \frac{- 8}{3} + 4 + \frac{1}{3} - 1 + 9 - 9 - \frac{8}{3} + 4\]

= 2

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पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 30 | पृष्ठ १२२

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