What is the integral from 0 to 1 of 7e^x-7xe^{x^2} ?

The integral from 0 to 1 of 7e^x-7xe^{x^2} is 7(e-1)-(7(e-1))/2

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integral from 0 to 1 of 7e^x-7xe^{x^2}

\int_{0}^{1} 7 e^{x} - 7 x e^{x^{2}} d x

7 (e - 1) - \frac{7 ( e - 1 )}{2}

Decimal

6.01398 \dots

Solution steps

\int_{0}^{1} 7 e^{x} - 7 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} 7 e^{x} d x - \int_{0}^{1} 7 x e^{x^{2}} d x

\int_{0}^{1} 7 e^{x} d x = 7 (e - 1)

\int_{0}^{1} 7 x e^{x^{2}} d x = \frac{7 ( e - 1 )}{2}

= 7 (e - 1) - \frac{7 ( e - 1 )}{2}

What is the integral from 0 to 1 of 7e^x-7xe^{x^2} ?
The integral from 0 to 1 of 7e^x-7xe^{x^2} is 7(e-1)-(7(e-1))/2