What is the integral from 0 to 1 of (3e^x-3xe^{x^2}) ?

The integral from 0 to 1 of (3e^x-3xe^{x^2}) is 3(e-1)-(3(e-1))/2

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integral from 0 to 1 of (3e^x-3xe^{x^2})

\int_{0}^{1} (3 e^{x} - 3 x e^{x^{2}}) d x

3 (e - 1) - \frac{3 ( e - 1 )}{2}

Decimal

2.57742 \dots

Solution steps

\int_{0}^{1} 3 e^{x} - 3 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} 3 e^{x} d x - \int_{0}^{1} 3 x e^{x^{2}} d x

\int_{0}^{1} 3 e^{x} d x = 3 (e - 1)

\int_{0}^{1} 3 x e^{x^{2}} d x = \frac{3 ( e - 1 )}{2}

= 3 (e - 1) - \frac{3 ( e - 1 )}{2}

What is the integral from 0 to 1 of (3e^x-3xe^{x^2}) ?
The integral from 0 to 1 of (3e^x-3xe^{x^2}) is 3(e-1)-(3(e-1))/2