What is the integral from 0.3 to 2.3 of 7xe^{-4x} ?

The integral from 0.3 to 2.3 of 7xe^{-4x} is 0.28944…

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integral from 0.3 to 2.3 of 7xe^{-4x}

\int_{0.3}^{2.3} 7 x e^{- 4 x} d x

0.28944 \dots

Solution steps

\int_{0.3}^{2.3} 7 x e^{- 4 x} d x

Take the constant out:

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

= 7 \cdot \int_{0.3}^{2.3} x e^{- 4 x} d x

Apply u-substitution

= 7 \cdot \int_{- 1.2}^{- 9.2} \frac{e ^{u} u}{16} d u

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

= 7 (- \int_{- 9.2}^{- 1.2} \frac{e ^{u} u}{16} d u)

Take the constant out:

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

= 7 (- \frac{1}{16} \cdot \int_{- 9.2}^{- 1.2} e^{u} u d u)

Apply Integration By Parts

= 7 (- \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 9.2}^{- 1.2})

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

= 7 (- \frac{1}{16} [e^{u} u - e^{u}]_{- 9.2}^{- 1.2})

Simplify

7 (- \frac{1}{16} [e^{u} u - e^{u}]_{- 9.2}^{- 1.2}) : - 0.4375 [e^{u} u - e^{u}]_{- 9.2}^{- 1.2}

= - 0.4375 [e^{u} u - e^{u}]_{- 9.2}^{- 1.2}

Compute the boundaries:

- 0.66159 \dots

= - 0.4375 (- 0.66159 \dots)

Simplify

= 0.28944 \dots

What is the integral from 0.3 to 2.3 of 7xe^{-4x} ?
The integral from 0.3 to 2.3 of 7xe^{-4x} is 0.28944…