What is the integral from 0 to 1 of 7/(1+p^2) ?

The integral from 0 to 1 of 7/(1+p^2) is (7pi)/4

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 7/(1+p^2)

\int_{0}^{1} \frac{7}{1 + p ^{2}} d p

\frac{7 π}{4}

Decimal

5.49778 \dots

Solution steps

\int_{0}^{1} \frac{7}{1 + p ^{2}} d p

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int_{0}^{1} \frac{1}{1 + p ^{2}} d p

Use the common integral:

\int \frac{1}{1 + p ^{2}} d p = arctan (p)

= 7 [arctan (p)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 7 \cdot \frac{π}{4}

Simplify

= \frac{7 π}{4}

\int_{0}^{1} sin (nπ t) d t

\int_{x^{3}}^{x} (e^{\frac{2 y}{x}}) d y

\int_{1}^{10} \frac{8}{x} d x

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

\int_{9}^{25} x x d x

What is the integral from 0 to 1 of 7/(1+p^2) ?
The integral from 0 to 1 of 7/(1+p^2) is (7pi)/4