Let $$[x]$$ denote the greatest integer less than or equal to $$x$$. The value of the integral $$\int_{0}^{\sqrt{2}}Â [x^2]e^x dx$$ is equal to
=$$\int_{0}^{1} 0 e^{x}dx+ \int_{1}^{\sqrt{2}} 1 e^{x} dx$$
=0 +Â Â $$e^{\sqrt{\ 2}}-e^1$$
=$$e^{\sqrt{2}} - e^{1}$$
Hence C is the correct answer.
Create a FREE account and get: