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

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}

\int e^{-}

\frac{d}{d x} (\frac{1}{2 π} e^{\frac{- x ^{2}}{2}})

x \to 3 lim (\frac{x - 3}{x + 3})

x y^{^{'}} + 4 y = - 3 x

\int - \frac{x}{x ^{2} + 4} 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)