What is the integral from 0 to 1 of (10)/(1+p^2) ?

The integral from 0 to 1 of (10)/(1+p^2) is (5pi)/2

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integral from 0 to 1 of (10)/(1+p^2)

\int_{0}^{1} \frac{10}{1 + p ^{2}} d p

\frac{5 π}{2}

Decimal

7.85398 \dots

Solution steps

\int_{0}^{1} \frac{10}{1 + p ^{2}} d p

Take the constant out:

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

= 10 \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)

= 10 [arctan (p)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 10 \cdot \frac{π}{4}

Simplify

= \frac{5 π}{2}

What is the integral from 0 to 1 of (10)/(1+p^2) ?
The integral from 0 to 1 of (10)/(1+p^2) is (5pi)/2