What is the integral from 0 to 1 of-8x5^x ?

The integral from 0 to 1 of-8x5^x is -(8(5ln(5)-4))/(ln^2(5))

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integral from 0 to 1 of-8x5^x

\int_{0}^{1} - 8 x 5^{x} d x

- \frac{8 ( 5 ln ( 5 ) - 4 )}{ln ^{2} ( 5 )}

Decimal Notation

- 12.49957 \dots

Solution steps

\int_{0}^{1} - 8 x \cdot 5^{x} d x

Take the constant out:

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

= - 8 \cdot \int_{0}^{1} x \cdot 5^{x} d x

Apply Integration By Parts

: [\frac{5 ^{x} x}{ln ( 5 )} - \int \frac{5 ^{x}}{ln ( 5 )} d x]_{0}^{1}

= - 8 [\frac{5 ^{x} x}{ln ( 5 )} - \int \frac{5 ^{x}}{ln ( 5 )} d x]_{0}^{1}

\int \frac{5 ^{x}}{ln ( 5 )} d x = \frac{5 ^{x}}{ln ^{2} ( 5 )}

= - 8 [\frac{5 ^{x} x}{ln ( 5 )} - \frac{5 ^{x}}{ln ^{2} ( 5 )}]_{0}^{1}

Compute the boundaries:

\frac{5 ln ( 5 ) - 4}{ln ^{2} ( 5 )}

= - 8 \cdot \frac{5 ln ( 5 ) - 4}{ln ^{2} ( 5 )}

Simplify

= - \frac{8 ( 5 ln ( 5 ) - 4 )}{ln ^{2} ( 5 )}

\frac{\partial}{\partial y} (e^{y} cos (x) - y^{2})

\int cos^{8} (θ) sin (θ) d θ

y - 3 = 5 (x - 2) y - 3 = 5 (x - 2)

\frac{\partial}{\partial x} (z^{2} y^{2} - e^{x})

\int \frac{26}{( 169 x ^{2} + 1 ) ^{2}} d x

What is the integral from 0 to 1 of-8x5^x ?
The integral from 0 to 1 of-8x5^x is -(8(5ln(5)-4))/(ln^2(5))