What is the integral from 0 to 2 of (4x-e^x) ?

The integral from 0 to 2 of (4x-e^x) is -e^2+9

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 (4x-e^x)

\int_{0}^{2} (4 x - e^{x}) d x

- e^{2} + 9

Decimal

1.61094 \dots

Solution steps

\int_{0}^{2} 4 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} 4 x d x - \int_{0}^{2} e^{x} d x

\int_{0}^{2} 4 x d x = 8

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

= 8 - (e^{2} - 1)

Simplify

= - e^{2} + 9

\int_{0}^{\frac{π}{4}} tan (y) d y

\int_{- 1}^{1} x^{\frac{4}{3}} + 4 x^{\frac{1}{3}} d x

\int_{0}^{2 π} sin (n θ) d θ

\int_{- 1}^{1} (a x^{2} + b x + c) d x

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

What is the integral from 0 to 2 of (4x-e^x) ?
The integral from 0 to 2 of (4x-e^x) is -e^2+9