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

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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

- 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}

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