Aspire Faculty ID #13872 · Topic: JAMIA MILLIA ISLAMIA MCA 2021 · Just now
JAMIA MILLIA ISLAMIA MCA 2021

Find the value of $I = \displaystyle\int_{-1}^{1} x^2 e^{[x]} dx$, where $[\,]$ denotes the greatest integer function.

Solution

From $-1$ to $0$: $[x] = -1$ From $0$ to $1$: $[x] = 0$ So, $I = \int_{-1}^{0} x^2 e^{-1} dx + \int_{0}^{1} x^2 e^0 dx$ $= e^{-1}\int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$ $= e^{-1}\left[\dfrac{x^3}{3}\right]{-1}^{0} + \left[\dfrac{x^3}{3}\right]{0}^{1}$ $= e^{-1}\left(\dfrac{1}{3}\right) + \dfrac{1}{3}$ $I = \dfrac{1}{3e} + \dfrac{1}{3}$

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