What is the integral from 0 to 1 of (4x^e+7e^x) ?

The integral from 0 to 1 of (4x^e+7e^x) is 4/(e+1)+7(e-1)

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 (4x^e+7e^x)

\int_{0}^{1} (4 x^{e} + 7 e^{x}) d x

\frac{4}{e + 1} + 7 (e - 1)

Decimal

13.10373 \dots

Solution steps

\int_{0}^{1} 4 x^{e} + 7 e^{x} 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} 4 x^{e} d x + \int_{0}^{1} 7 e^{x} d x

\int_{0}^{1} 4 x^{e} d x = \frac{4}{e + 1}

\int_{0}^{1} 7 e^{x} d x = 7 (e - 1)

= \frac{4}{e + 1} + 7 (e - 1)

\int_{0}^{t} (t - u) y (u) d u

\int_{0}^{e} \frac{x ^{3}}{1 + x ^{2}} d x

\int_{0}^{1} (\frac{18 x}{x ^{2} + 1}) d x

\int_{0}^{4} 48 - 3 x^{2} d x

\int_{1}^{3} (9 - x^{2}) d x

What is the integral from 0 to 1 of (4x^e+7e^x) ?
The integral from 0 to 1 of (4x^e+7e^x) is 4/(e+1)+7(e-1)