What is the integral from 0 to 1 of xe^{7x} ?

The integral from 0 to 1 of xe^{7x} is (6e^7+1)/(49)

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integral from 0 to 1 of xe^{7x}

\int_{0}^{1} x e^{7 x} d x

\frac{6 e ^{7} + 1}{49}

Decimal Notation

134.30201 \dots

Solution steps

\int_{0}^{1} x e^{7 x} d x

Apply u-substitution

: \int_{0}^{7} \frac{e ^{u} u}{49} d u

= \int_{0}^{7} \frac{e ^{u} u}{49} d u

Take the constant out:

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

= \frac{1}{49} \cdot \int_{0}^{7} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{7}

= \frac{1}{49} [e^{u} u - \int e^{u} d u]_{0}^{7}

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

= \frac{1}{49} [e^{u} u - e^{u}]_{0}^{7}

Compute the boundaries:

6 e^{7} + 1

= \frac{1}{49} (6 e^{7} + 1)

Simplify

= \frac{6 e ^{7} + 1}{49}

y = \frac{5}{x} (1.5)

y^{^{''}} + 14 y^{^{'}} + 65 y = 0

f (x) = 3 x + 10, (2, 4)

\int \frac{26}{( x - 1 ) ( x ^{2} + 25 )} d x

(x^{2} - 7 x + 4) 25 - x^{2}

What is the integral from 0 to 1 of xe^{7x} ?
The integral from 0 to 1 of xe^{7x} is (6e^7+1)/(49)