What is the integral from 0 to 3 of rsqrt(1+4r^2) ?

The integral from 0 to 3 of rsqrt(1+4r^2) is (37sqrt(37)-1)/(12)

Popular >

integral from 0 to 3 of rsqrt(1+4r^2)

\int_{0}^{3} r 1 + 4 r^{2} d r

\frac{37 37 - 1}{12}

Decimal

18.67185 \dots

Solution steps

\int_{0}^{3} r 1 + 4 r^{2} d r

Apply u-substitution

: \int_{1}^{37} \frac{u}{8} d u

= \int_{1}^{37} \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{37} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

Compute the boundaries:

\frac{74 37}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{74 37}{3} - \frac{2}{3})

Simplify

= \frac{37 37 - 1}{12}

\int_{2}^{10} 1

\int_{- 1}^{1} x \cdot \frac{3}{2} x^{2} d x

\int_{1}^{e^{2}} \frac{1}{x ( ln ( x ) ) ^{3}} d x

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

\int_{0}^{7} x^{2} - 6 x + 5 d x

What is the integral from 0 to 3 of rsqrt(1+4r^2) ?
The integral from 0 to 3 of rsqrt(1+4r^2) is (37sqrt(37)-1)/(12)