What is the integral from 1 to 4 of tsqrt(9+16t^2) ?

The integral from 1 to 4 of tsqrt(9+16t^2) is (5(53sqrt(265)-25))/(48)

Popular >

integral from 1 to 4 of tsqrt(9+16t^2)

\int_{1}^{4} t 9 + 16 t^{2} d t

\frac{5 ( 53 265 - 25 )}{48}

Decimal

87.26848 \dots

Solution steps

\int_{1}^{4} t 9 + 16 t^{2} d t

Apply u-substitution

: \int_{25}^{265} \frac{u}{32} d u

= \int_{25}^{265} \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}^{265} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{265}

= \frac{1}{32} [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{265}

Compute the boundaries:

\frac{530 265}{3} - \frac{250}{3}

= \frac{1}{32} (\frac{530 265}{3} - \frac{250}{3})

Simplify

= \frac{5 ( 53 265 - 25 )}{48}

\int_{0}^{2} (2 x^{3} + 2 x) d x

\int_{1}^{11} \frac{3 ( ln ( x ) ) ^{2}}{x} d x

\int_{0}^{10} 5 x d x

\int_{0}^{9} 9 - x d x

\int_{\frac{π}{4}}^{π} (sin (θ)) d θ

What is the integral from 1 to 4 of tsqrt(9+16t^2) ?
The integral from 1 to 4 of tsqrt(9+16t^2) is (5(53sqrt(265)-25))/(48)