The value of ∫ - 1 1 x 2 e x 3 d x , where t denotes the greatest integer ≤ t , is :

I = ∫ - 1 1 x 2 e x 3 d x = ∫ - 1 0 x 2 e x 3 d x + ∫ 0 1 x 2 e x 3 d x = ∫ - 1 0 x 2 e - 1 d x + ∫ 0 1 x 2 e 0 d x = 1 e × x 3 3 0 - 1 + x 3 3 0 1 = 1 e × 0 - - 1 3 + 1 3 = 1 3 e + 1 3 = 1 + e 3 e

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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationProperties of definite integrationHard2 minPYQ_2021

MathematicsHardsingle choice

The value of $\int_{- 1}^{1} x^{2} e^{[x^{3}]} d x,$ where $[t]$ denotes the greatest integer $\leq t,$ is :

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Answer:

Solution:

$I = \int_{- 1}^{1} x^{2} e^{[x^{3}]} d x$

$= \int_{- 1}^{0} x^{2} e^{[x^{3}]} d x + \int_{0}^{1} x^{2} e^{[x^{3}]} d x$

$= \int_{- 1}^{0} x^{2} e^{- 1} d x + \int_{0}^{1} x^{2} e^{0} d x$

$= \frac{1}{e} \times \frac{x^{3}}{3} {|\begin{matrix} 0 \\ - 1 \end{matrix} + \frac{x^{3}}{3}|}_{0}^{1}$

$= \frac{1}{e} \times (0 - (\frac{- 1}{3})) + \frac{1}{3}$

$= \frac{1}{3 e} + \frac{1}{3} = \frac{1 + e}{3 e}$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Properties of definite integration

⏱ 2mℹ️ Source: PYQ_2021

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