What is the integral from 0 to 1 of (7x^e+8e^x) ?

The integral from 0 to 1 of (7x^e+8e^x) is 7/(e+1)+8(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 (7x^e+8e^x)

\int_{0}^{1} (7 x^{e} + 8 e^{x}) d x

\frac{7}{e + 1} + 8 (e - 1)

Decimal

15.62884 \dots

Solution steps

\int_{0}^{1} 7 x^{e} + 8 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} 7 x^{e} d x + \int_{0}^{1} 8 e^{x} d x

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

\int_{0}^{1} 8 e^{x} d x = 8 (e - 1)

= \frac{7}{e + 1} + 8 (e - 1)

\int_{0}^{2 π} 9 cos^{2} (t) d t

\int_{3}^{\infty} \frac{1}{9 + x ^{2}} d x

\int_{0}^{1} \frac{e ^{x} + 9}{e ^{x} + 9 x} d x

\int_{0}^{1} 10 x (2 x^{2} + 1)^{4} d x

\int_{- 1}^{2} (3 x - 5) (x + 2) d x

What is the integral from 0 to 1 of (7x^e+8e^x) ?
The integral from 0 to 1 of (7x^e+8e^x) is 7/(e+1)+8(e-1)