What is the integral from 0 to 2 of sqrt(36t^2+9t^4) ?

The integral from 0 to 2 of sqrt(36t^2+9t^4) is 16sqrt(2)-8

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integral from 0 to 2 of sqrt(36t^2+9t^4)

\int_{0}^{2} 36 t^{2} + 9 t^{4} d t

162 - 8

Decimal

14.62741 \dots

Solution steps

\int_{0}^{2} 36 t^{2} + 9 t^{4} d t

Factor

36 t^{2} + 9 t^{4} : 3 t^{2} 4 + t^{2}

= \int_{0}^{2} 3 t^{2} 4 + t^{2} d t

Take the constant out:

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

= 3 \cdot \int_{0}^{2} t^{2} 4 + t^{2} d t

t^{2} = (t),

assuming

t \geq 0

= 3 \cdot \int_{0}^{2} t 4 + t^{2} d t

Apply u-substitution

: \int_{2}^{22} u^{2} d u

= 3 \cdot \int_{2}^{22} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{2}^{22}

= 3 [\frac{u ^{3}}{3}]_{2}^{22}

Compute the boundaries:

\frac{16 2}{3} - \frac{8}{3}

= 3 (\frac{16 2}{3} - \frac{8}{3})

Simplify

= 162 - 8

\int_{0}^{\infty} 1 - e^{- 2 x} d x

\int_{- x}^{x} sin (x) d x

\int_{- \infty}^{0} \frac{2}{x ^{2} + 4} d x

\int_{- 1}^{2} (x^{2} + 1) (x - 3) d x

\int_{0}^{2} (2 x - 9) (4 x^{2} + 8) d x

What is the integral from 0 to 2 of sqrt(36t^2+9t^4) ?
The integral from 0 to 2 of sqrt(36t^2+9t^4) is 16sqrt(2)-8