What is the integral from 1 to 3 of (8x-1)e^{4x^2-x} ?

The integral from 1 to 3 of (8x-1)e^{4x^2-x} is e^{33}-e^3

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integral from 1 to 3 of (8x-1)e^{4x^2-x}

\int_{1}^{3} (8 x - 1) e^{4 x^{2} - x} d x

e^{33} - e^{3}

Decimal

2.14644 E 14

Solution steps

\int_{1}^{3} (8 x - 1) e^{4 x^{2} - x} d x

Apply u-substitution

: \int_{3}^{33} e^{u} d u

= \int_{3}^{33} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= [e^{u}]_{3}^{33}

Compute the boundaries:

e^{33} - e^{3}

= e^{33} - e^{3}

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What is the integral from 1 to 3 of (8x-1)e^{4x^2-x} ?
The integral from 1 to 3 of (8x-1)e^{4x^2-x} is e^{33}-e^3