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

The integral from 0 to 4 of sqrt(4t^2+9t^4) is (8(37sqrt(37)-1))/(27)

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

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

\frac{8 ( 37 37 - 1 )}{27}

Decimal Notation

66.38880 \dots

Solution steps

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

Factor

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

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

t^{2} = (t),

assuming

t \geq 0

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

Apply u-substitution

: \int_{4}^{148} \frac{u}{18} d u

= \int_{4}^{148} \frac{u}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int_{4}^{148} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{148}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{148}

Compute the boundaries:

\frac{592 37}{3} - \frac{16}{3}

= \frac{1}{18} (\frac{592 37}{3} - \frac{16}{3})

Simplify

= \frac{8 ( 37 37 - 1 )}{27}

x \to 5 lim (csc (\frac{π x}{4}))

\frac{\partial}{\partial x} (x \cdot sin (2 y))

\frac{d}{d y} (x^{2} - y)

\int \frac{x ^{2} + x + 3}{x - 2} d x

y = x^{2} - 2, y = \frac{( 3 x + 1 )}{2}

What is the integral from 0 to 4 of sqrt(4t^2+9t^4) ?
The integral from 0 to 4 of sqrt(4t^2+9t^4) is (8(37sqrt(37)-1))/(27)