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))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

- 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

= - 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 )}

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))