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

The integral from 0 to 1 of 7/(1+t^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+t^2)

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

\frac{7 π}{4}

Decimal Notation

5.49778 \dots

Solution steps

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

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 + t ^{2}} d t

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{4}

= 7 \cdot \frac{π}{4}

Simplify

= \frac{7 π}{4}

\frac{d}{d x} (6 + 2 ln (x))

\frac{d}{d x} (x - 5 x^{2})

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

\frac{d x}{d t} - 3 x = cos (4 t)

x \to 2 - lim (\frac{4}{( x - 2 ) ^{3}})

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