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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{98}{3}

Decimal

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_{0}^{5} (3 t - 8) d t

\int_{2}^{3} (3 x + 8) d x

\int_{0}^{1} 30 x y d y

\int_{- 2}^{1} (x^{2}) d x

\int_{1}^{4} 2 π x (1 + x) 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