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

The integral from 0 to 1 of (x^2)/(1+y^2) is (pix^2)/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 (x^2)/(1+y^2)

\int_{0}^{1} \frac{x ^{2}}{1 + y ^{2}} d y

\frac{π x ^{2}}{4}

Solution steps

\int_{0}^{1} \frac{x ^{2}}{1 + y ^{2}} d y

Take the constant out:

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

= x^{2} \cdot \int_{0}^{1} \frac{1}{1 + y ^{2}} d y

Use the common integral:

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

= x^{2} [arctan (y)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= x^{2} \frac{π}{4}

Simplify

= \frac{π x ^{2}}{4}

\int_{0}^{4} (2 x + 5) (3 x - 1) d x

\int_{0}^{8} x^{\frac{7}{3}} d x

\int_{1}^{2} x^{l n (x)} d x

\int_{2}^{10} 2 x^{2} d x

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

What is the integral from 0 to 1 of (x^2)/(1+y^2) ?
The integral from 0 to 1 of (x^2)/(1+y^2) is (pix^2)/4