What is the integral from 1 to 2 of 12(y-1)x^3e^{2x(y-1)} ?

The integral from 1 to 2 of 12(y-1)x^3e^{2x(y-1)} is 3x(2xe^{2x}-e^{2x}+1)

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integral from 1 to 2 of 12(y-1)x^3e^{2x(y-1)}

\int_{1}^{2} 12 (y - 1) x^{3} e^{2 x (y - 1)} d y

3 x (2 x e^{2 x} - e^{2 x} + 1)

Solution steps

\int_{1}^{2} 12 (y - 1) x^{3} e^{2 x (y - 1)} d y

Take the constant out:

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

= 12 x^{3} \cdot \int_{1}^{2} (- 1 + y) e^{(- 1 + y) \cdot 2 x} d y

Apply u-substitution

: \int_{0}^{1} e^{2 xu} u d u

= 12 x^{3} \cdot \int_{0}^{1} e^{2 xu} u d u

Apply Integration By Parts

: [\frac{1}{2} e^{2 xu} u^{2} - \int x e^{2 xu} u^{2} d u]_{0}^{1}

= 12 x^{3} [\frac{1}{2} e^{2 xu} u^{2} - \int x e^{2 xu} u^{2} d u]_{0}^{1}

\int x e^{2 xu} u^{2} d u = \frac{e ^{2 xu} ( 2 x ^{2} u ^{2} - 2 xu + 1 )}{4 x ^{2}}

= 12 x^{3} [\frac{1}{2} e^{2 xu} u^{2} - \frac{e ^{2 xu} ( 2 x ^{2} u ^{2} - 2 xu + 1 )}{4 x ^{2}}]_{0}^{1}

Join

\frac{1}{2} e^{2 xu} u^{2} - \frac{e ^{2 xu} ( 2 x ^{2} u ^{2} - 2 xu + 1 )}{4 x ^{2}} : \frac{2 x e ^{2 xu} u - e ^{2 xu}}{4 x ^{2}}

= 12 x^{3} [\frac{2 x e ^{2 xu} u - e ^{2 xu}}{4 x ^{2}}]_{0}^{1}

Compute the boundaries:

\frac{2 x e ^{2 x} - e ^{2 x}}{4 x ^{2}} + \frac{1}{4 x ^{2}}

= 12 x^{3} (\frac{2 x e ^{2 x} - e ^{2 x}}{4 x ^{2}} + \frac{1}{4 x ^{2}})

Simplify

= 3 x (2 x e^{2 x} - e^{2 x} + 1)

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What is the integral from 1 to 2 of 12(y-1)x^3e^{2x(y-1)} ?
The integral from 1 to 2 of 12(y-1)x^3e^{2x(y-1)} is 3x(2xe^{2x}-e^{2x}+1)