What is the integral from 1 to 3 of 1/2 (3^2-1^2) ?

The integral from 1 to 3 of 1/2 (3^2-1^2) is 8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 3 of 1/2 (3^2-1^2)

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

8

Solution steps

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

Integral of a constant:

\int a d x = a x

= [\frac{1}{2} (3^{2} - 1^{2}) x]_{1}^{3}

Simplify

[\frac{1}{2} (3^{2} - 1^{2}) x]_{1}^{3} : [4 x]_{1}^{3}

= [4 x]_{1}^{3}

Compute the boundaries:

8

= 8

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

\int_{0}^{\frac{π}{4}} \frac{3 cos ^{2} ( θ )}{2} d θ

\int_{2}^{6} \frac{2}{x} - \frac{4}{x ^{2}} d x

\int_{0}^{5} \frac{10}{x} d x

\int_{0}^{6} (x^{2} + 2) d x

What is the integral from 1 to 3 of 1/2 (3^2-1^2) ?
The integral from 1 to 3 of 1/2 (3^2-1^2) is 8