What is the integral from 0 to 4 of xsqrt(x^2+9) ?

The integral from 0 to 4 of xsqrt(x^2+9) is 98/3

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integral from 0 to 4 of xsqrt(x^2+9)

\int_{0}^{4} x x^{2} + 9 d x

\frac{98}{3}

Decimal Notation

32.66666 \dots

Solution steps

\int_{0}^{4} x x^{2} + 9 d x

Apply u-substitution

: \int_{3}^{5} u^{2} d u

= \int_{3}^{5} u^{2} d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{98}{3}

= \frac{98}{3}

\int \frac{1}{x ^{4} x ^{2} + 3} d x

x \to 0 lim (\frac{x ^{3}}{x ^{4}})

y = 5 x^{2}, (- 2, 20)

\frac{d}{d x} (f (x) (x^{2}))

\int \frac{5 - 50 x}{1 - 25 x ^{2}} d x

What is the integral from 0 to 4 of xsqrt(x^2+9) ?
The integral from 0 to 4 of xsqrt(x^2+9) is 98/3