What is the integral from 0 to 1 of-5x5^x ?

The integral from 0 to 1 of-5x5^x is -(5(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

Popular >

integral from 0 to 1 of-5x5^x

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

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

Decimal Notation

- 7.81223 \dots

Solution steps

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

Take the constant out:

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

= - 5 \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}

= - 5 [\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 )}

= - 5 [\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 )}

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

Simplify

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

\int \frac{3}{( x - 2 ) ( x ^{2} + 4 )} d x

\frac{d}{d x} (\frac{3 x}{1 - x})

\int \frac{( x ^{\frac{1}{3}} )}{x ^{\frac{2}{3}}} d x

\int \frac{1}{( 8 - 2 x - x ^{2} ) ^{\frac{3}{2}}} d x

n = 0 \sum \infty \frac{2 ^{n}}{4 ^{n}}

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