What is the integral from 0 to 1 of (tsqrt(4+9t^2)) ?

The integral from 0 to 1 of (tsqrt(4+9t^2)) is (13sqrt(13)-8)/(27)

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integral from 0 to 1 of (tsqrt(4+9t^2))

\int_{0}^{1} (t 4 + 9 t^{2}) d t

\frac{13 13 - 8}{27}

Decimal Notation

1.43970 \dots

Solution steps

\int_{0}^{1} t 4 + 9 t^{2} d t

Apply u-substitution

: \int_{4}^{13} \frac{u}{18} d u

= \int_{4}^{13} \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}^{13} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{16}{3}

= \frac{1}{18} (\frac{26 13}{3} - \frac{16}{3})

Simplify

= \frac{13 13 - 8}{27}

x + 2 y = x + 2 y = 8

y^{^{'}} = 5 x (x - 3)

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

\frac{\partial}{\partial x} (e^{y e^{x}})

\int (e^{- 5 x} + 2 e^{x}) d x

What is the integral from 0 to 1 of (tsqrt(4+9t^2)) ?
The integral from 0 to 1 of (tsqrt(4+9t^2)) is (13sqrt(13)-8)/(27)