What is the integral from 0 to 1 of tsqrt(4+5t) ?

The integral from 0 to 1 of tsqrt(4+5t) is 506/375

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integral from 0 to 1 of tsqrt(4+5t)

\int_{0}^{1} t 4 + 5 t d t

\frac{506}{375}

Decimal

1.34933 \dots

Solution steps

\int_{0}^{1} t 4 + 5 t d t

Apply u-substitution

: \int_{4}^{9} \frac{u ( u - 4 )}{25} d u

= \int_{4}^{9} \frac{u ( u - 4 )}{25} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int_{4}^{9} u (u - 4) d u

Expand

u (u - 4) : u^{\frac{3}{2}} - 4 u

= \frac{1}{25} \cdot \int_{4}^{9} u^{\frac{3}{2}} - 4 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{25} (\int_{4}^{9} u^{\frac{3}{2}} d u - \int_{4}^{9} 4 u d u)

\int_{4}^{9} u^{\frac{3}{2}} d u = \frac{422}{5}

\int_{4}^{9} 4 u d u = \frac{152}{3}

= \frac{1}{25} (\frac{422}{5} - \frac{152}{3})

Simplify

\frac{1}{25} (\frac{422}{5} - \frac{152}{3}) : \frac{506}{375}

= \frac{506}{375}

\int_{- 2}^{1} x^{3} + 2 x^{2} - x + 2 d x

\int_{1}^{4} \frac{4}{x} - \frac{x}{4} d x

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

\int_{0}^{2} 2 - x^{3} d x

\int_{2}^{3} \frac{2}{x ^{2}} d x

What is the integral from 0 to 1 of tsqrt(4+5t) ?
The integral from 0 to 1 of tsqrt(4+5t) is 506/375