What is the integral from 0 to 1 of 5e^x-5xe^{x^2} ?

The integral from 0 to 1 of 5e^x-5xe^{x^2} is 5(e-1)-(5(e-1))/2

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integral from 0 to 1 of 5e^x-5xe^{x^2}

\int_{0}^{1} 5 e^{x} - 5 x e^{x^{2}} d x

5 (e - 1) - \frac{5 ( e - 1 )}{2}

Decimal

4.29570 \dots

Solution steps

\int_{0}^{1} 5 e^{x} - 5 x e^{x^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 5 e^{x} d x - \int_{0}^{1} 5 x e^{x^{2}} d x

\int_{0}^{1} 5 e^{x} d x = 5 (e - 1)

\int_{0}^{1} 5 x e^{x^{2}} d x = \frac{5 ( e - 1 )}{2}

= 5 (e - 1) - \frac{5 ( e - 1 )}{2}

What is the integral from 0 to 1 of 5e^x-5xe^{x^2} ?
The integral from 0 to 1 of 5e^x-5xe^{x^2} is 5(e-1)-(5(e-1))/2