What is the integral from 6 to 0 of (2+5x)e^{1/3 x} ?

The integral from 6 to 0 of (2+5x)e^{1/3 x} is -51e^2-39

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integral from 6 to 0 of (2+5x)e^{1/3 x}

\int_{6}^{0} (2 + 5 x) e^{\frac{1}{3} x} d x

- 51 e^{2} - 39

Decimal

- 415.84186 \dots

Solution steps

\int_{6}^{0} (2 + 5 x) e^{\frac{1}{3} x} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{6} (2 + 5 x) e^{\frac{1}{3} x} d x

Expand

(2 + 5 x) e^{\frac{1}{3} x} : 2 e^{\frac{x}{3}} + 5 e^{\frac{x}{3}} x

= - \int_{0}^{6} 2 e^{\frac{x}{3}} + 5 e^{\frac{x}{3}} x 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}^{6} 2 e^{\frac{x}{3}} d x + \int_{0}^{6} 5 e^{\frac{x}{3}} x d x)

\int_{0}^{6} 2 e^{\frac{x}{3}} d x = 6 (e^{2} - 1)

\int_{0}^{6} 5 e^{\frac{x}{3}} x d x = 5 (9 e^{2} + 9)

= - (6 (e^{2} - 1) + 5 (9 e^{2} + 9))

Simplify

- (6 (e^{2} - 1) + 5 (9 e^{2} + 9)) : - 51 e^{2} - 39

= - 51 e^{2} - 39

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What is the integral from 6 to 0 of (2+5x)e^{1/3 x} ?
The integral from 6 to 0 of (2+5x)e^{1/3 x} is -51e^2-39