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

The integral from 0 to 3 of xsqrt(x^2+16) is 61/3

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

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

\frac{61}{3}

Decimal Notation

20.33333 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{61}{3}

= \frac{61}{3}

\frac{d}{d x} (x^{2} sin (3 x - 2))

\int_{0}^{1} \frac{1}{x} d x

\frac{d}{d x} (sec^{2} (x) - tan^{2} (x))

t \to 0 lim (\frac{5 ^{t} - 3 ^{t}}{t})

f^{^{'}} (a) = 2

What is the integral from 0 to 3 of xsqrt(x^2+16) ?
The integral from 0 to 3 of xsqrt(x^2+16) is 61/3