What is the integral from 0 to 3 of 2xsqrt(x^2+16) ?

The integral from 0 to 3 of 2xsqrt(x^2+16) is 122/3

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integral from 0 to 3 of 2xsqrt(x^2+16)

\int_{0}^{3} 2 x x^{2} + 16 d x

\frac{122}{3}

Decimal

40.66666 \dots

Solution steps

\int_{0}^{3} 2 x x^{2} + 16 d x

Take the constant out:

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

= 2 \cdot \int_{0}^{3} x x^{2} + 16 d x

Apply u-substitution

: \int_{4}^{5} u^{2} d u

= 2 \cdot \int_{4}^{5} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{4}^{5}

= 2 [\frac{u ^{3}}{3}]_{4}^{5}

Compute the boundaries:

\frac{61}{3}

= 2 \cdot \frac{61}{3}

Simplify

= \frac{122}{3}

\int_{4}^{5} x (x^{2} - 16)^{3} d x

\int_{0}^{3} 4 x - x^{2} d x

\int_{2}^{4} (x^{2} + 2) d x

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

\int_{1}^{3} \frac{1}{32} x y d y

What is the integral from 0 to 3 of 2xsqrt(x^2+16) ?
The integral from 0 to 3 of 2xsqrt(x^2+16) is 122/3