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

The integral from 0 to 1 of (x^2+6)e^{5x} is (167e^5-152)/(125)

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

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

\frac{167 e ^{5} - 152}{125}

Decimal Notation

197.06398 \dots

Solution steps

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

Expand

(x^{2} + 6) e^{5 x} : e^{5 x} x^{2} + 6 e^{5 x}

= \int_{0}^{1} e^{5 x} x^{2} + 6 e^{5 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}^{1} e^{5 x} x^{2} d x + \int_{0}^{1} 6 e^{5 x} d x

\int_{0}^{1} e^{5 x} x^{2} d x = \frac{17 e ^{5} - 2}{125}

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

= \frac{17 e ^{5} - 2}{125} + \frac{6 ( e ^{5} - 1 )}{5}

Simplify

\frac{17 e ^{5} - 2}{125} + \frac{6 ( e ^{5} - 1 )}{5} : \frac{167 e ^{5} - 152}{125}

= \frac{167 e ^{5} - 152}{125}

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\int_{0}^{4} (1 - e^{- 2 x}) d x

What is the integral from 0 to 1 of (x^2+6)e^{5x} ?
The integral from 0 to 1 of (x^2+6)e^{5x} is (167e^5-152)/(125)