Question 25

Let $$[x]$$ denote the greatest integer less than or equal to $$x$$. The value of the integral $$\int_{0}^{\sqrt{2}} [x^2]e^x dx$$ is equal to

Solution

=$$\int_{0}^{1} 0 e^{x}dx+ \int_{1}^{\sqrt{2}} 1 e^{x} dx$$

=0 + $$e^{\sqrt{\ 2}}-e^1$$

=$$e^{\sqrt{2}} - e^{1}$$

Hence C is the correct answer.

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