What is the integral from 0 to 1 of e^{7r-1}(2r^2-r) ?

The integral from 0 to 1 of e^{7r-1}(2r^2-r) is (32e^7-11)/(343e)

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integral from 0 to 1 of e^{7r-1}(2r^2-r)

\int_{0}^{1} e^{7 r - 1} (2 r^{2} - r) d r

\frac{32 e ^{7} - 11}{343 e}

Decimal Notation

37.62587 \dots

Solution steps

\int_{0}^{1} e^{7 r - 1} (2 r^{2} - r) d r

Apply u-substitution

: \int_{- 1}^{6} \frac{e ^{u} ( 2 u ^{2} - 3 u - 5 )}{343} d u

= \int_{- 1}^{6} \frac{e ^{u} ( 2 u ^{2} - 3 u - 5 )}{343} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{343} \cdot \int_{- 1}^{6} e^{u} (2 u^{2} - 3 u - 5) d u

Expand

e^{u} (2 u^{2} - 3 u - 5) : 2 e^{u} u^{2} - 3 e^{u} u - 5 e^{u}

= \frac{1}{343} \cdot \int_{- 1}^{6} 2 e^{u} u^{2} - 3 e^{u} u - 5 e^{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{343} (\int_{- 1}^{6} 2 e^{u} u^{2} d u - \int_{- 1}^{6} 3 e^{u} u d u - \int_{- 1}^{6} 5 e^{u} d u)

\int_{- 1}^{6} 2 e^{u} u^{2} d u = 2 (26 e^{6} - \frac{5}{e})

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

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

= \frac{1}{343} (2 (26 e^{6} - \frac{5}{e}) - 3 (5 e^{6} + \frac{2}{e}) - 5 (e^{6} - \frac{1}{e}))

Simplify

\frac{1}{343} (2 (26 e^{6} - \frac{5}{e}) - 3 (5 e^{6} + \frac{2}{e}) - 5 (e^{6} - \frac{1}{e})) : \frac{32 e ^{7} - 11}{343 e}

= \frac{32 e ^{7} - 11}{343 e}

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What is the integral from 0 to 1 of e^{7r-1}(2r^2-r) ?
The integral from 0 to 1 of e^{7r-1}(2r^2-r) is (32e^7-11)/(343e)