What is the integral from 0 to-1 of 3/(1+t^2) ?

The integral from 0 to-1 of 3/(1+t^2) is -(3pi)/4

Popular >

integral from 0 to-1 of 3/(1+t^2)

\int_{0}^{- 1} \frac{3}{1 + t ^{2}} d t

- \frac{3 π}{4}

Decimal Notation

- 2.35619 \dots

Solution steps

\int_{0}^{- 1} \frac{3}{1 + t ^{2}} d t

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{0} \frac{3}{1 + t ^{2}} d t

Take the constant out:

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

= - 3 \cdot \int_{- 1}^{0} \frac{1}{1 + t ^{2}} d t

Use the common integral:

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

= - 3 [arctan (t)]_{- 1}^{0}

Compute the boundaries:

\frac{π}{4}

= - 3 \cdot \frac{π}{4}

Simplify

= - \frac{3 π}{4}

\int_{1}^{e^{x}} 3 ln (x) d x

\int_{1}^{4} e^{- x} d x

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

\int_{- 2}^{2} - x^{2} + 5 d x

\int_{0}^{1} \frac{1}{2} e^{x} + \frac{3}{4} 2^{x} d x

What is the integral from 0 to-1 of 3/(1+t^2) ?
The integral from 0 to-1 of 3/(1+t^2) is -(3pi)/4