What is the integral from 0 to 2 of sin(x)+e^x ?

The integral from 0 to 2 of sin(x)+e^x is e^2-cos(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 2 of sin(x)+e^x

\int_{0}^{2} sin (x) + e^{x} d x

e^{2} - cos (2)

Decimal Notation

7.80520 \dots

Solution steps

\int_{0}^{2} sin (x) + 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}^{2} sin (x) d x + \int_{0}^{2} e^{x} d x

\int_{0}^{2} sin (x) d x = - cos (2) + 1

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

= - cos (2) + 1 + e^{2} - 1

Simplify

= e^{2} - cos (2)

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

s^{2} - 4 s + 3

\int (x^{4} - 1) x^{2} d x

\frac{d y}{d x} = y (2 - 3 y)

f (x) = ln (x^{2} + 2)

What is the integral from 0 to 2 of sin(x)+e^x ?
The integral from 0 to 2 of sin(x)+e^x is e^2-cos(2)