What is the integral from 1 to 2 of tsqrt(9+16t^2) ?

The integral from 1 to 2 of tsqrt(9+16t^2) is (73sqrt(73)-125)/(48)

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integral from 1 to 2 of tsqrt(9+16t^2)

\int_{1}^{2} t 9 + 16 t^{2} d t

\frac{73 73 - 125}{48}

Solution steps

\int_{1}^{2} t 9 + 16 t^{2} d t

Apply u-substitution

: \int_{25}^{73} \frac{u}{32} d u

= \int_{25}^{73} \frac{u}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int_{25}^{73} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{73}

= \frac{1}{32} [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{73}

Compute the boundaries:

\frac{146 73}{3} - \frac{250}{3}

= \frac{1}{32} (\frac{146 73}{3} - \frac{250}{3})

Simplify

= \frac{73 73 - 125}{48}

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What is the integral from 1 to 2 of tsqrt(9+16t^2) ?
The integral from 1 to 2 of tsqrt(9+16t^2) is (73sqrt(73)-125)/(48)