What is the integral from 1 to 4 of xe^{-2x^2}+6x-2 ?

The integral from 1 to 4 of xe^{-2x^2}+6x-2 is (1/(e^2)-1/(e^{32)})/4+39

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integral from 1 to 4 of xe^{-2x^2}+6x-2

\int_{1}^{4} x e^{- 2 x^{2}} + 6 x - 2 d x

\frac{\frac{1}{e ^{2}} - \frac{1}{e ^{32}}}{4} + 39

Decimal

39.03383 \dots

Solution steps

\int_{1}^{4} x e^{- 2 x^{2}} + 6 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_{1}^{4} x e^{- 2 x^{2}} d x + \int_{1}^{4} 6 x d x - \int_{1}^{4} 2 d x

\int_{1}^{4} x e^{- 2 x^{2}} d x = \frac{\frac{1}{e ^{2}} - \frac{1}{e ^{32}}}{4}

\int_{1}^{4} 6 x d x = 45

\int_{1}^{4} 2 d x = 6

= \frac{\frac{1}{e ^{2}} - \frac{1}{e ^{32}}}{4} + 45 - 6

Simplify

= \frac{\frac{1}{e ^{2}} - \frac{1}{e ^{32}}}{4} + 39

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What is the integral from 1 to 4 of xe^{-2x^2}+6x-2 ?
The integral from 1 to 4 of xe^{-2x^2}+6x-2 is (1/(e^2)-1/(e^{32)})/4+39