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प्रश्न

\[\int_{- 2}^2 x e^\left| x \right| dx\]
बेरीज
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उत्तर

Consider

\[f\left( x \right) = x e^\left| x \right|\]

Now,

\[f\left( - x \right) = \left( - x \right) e^\left| - x \right| = - x e^\left| x \right| = - f\left( x \right)\]

⇒ f(x) is an odd function.

\[\therefore \int_{- 2}^2 x e^\left| x \right| dx = 0\]

\[\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]

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पाठ 19: Definite Integrals - Exercise 20.3 [पृष्ठ ५६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.3 | Q 21 | पृष्ठ ५६

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