What is the integral from 0 to 1 of 5e^x-5xe^{x^2} ?

The integral from 0 to 1 of 5e^x-5xe^{x^2} is 5(e-1)-(5(e-1))/2

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 5e^x-5xe^{x^2}

\int_{0}^{1} 5 e^{x} - 5 x e^{x^{2}} d x

5 (e - 1) - \frac{5 ( e - 1 )}{2}

Decimal Notation

4.29570 \dots

Solution steps

\int_{0}^{1} 5 e^{x} - 5 x e^{x^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 5 e^{x} d x - \int_{0}^{1} 5 x e^{x^{2}} d x

\int_{0}^{1} 5 e^{x} d x = 5 (e - 1)

\int_{0}^{1} 5 x e^{x^{2}} d x = \frac{5 ( e - 1 )}{2}

= 5 (e - 1) - \frac{5 ( e - 1 )}{2}

\int \frac{16}{( x ^{2} + 4 ) ^{2}} d x

y = - 3 x - 9, (- 9, 18)

g (x) = 2 x^{0.8} + 4 x^{- 0.8}

\int 3 x^{2} + 2 y^{2} d x

f (x) = cot^{2} (cos (x))

What is the integral from 0 to 1 of 5e^x-5xe^{x^2} ?
The integral from 0 to 1 of 5e^x-5xe^{x^2} is 5(e-1)-(5(e-1))/2