What is the integral from 4 to 9 of (t-3)/(sqrt(t)) ?

The integral from 4 to 9 of (t-3)/(sqrt(t)) is 20/3

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integral from 4 to 9 of (t-3)/(sqrt(t))

\int_{4}^{9} \frac{t - 3}{t} d t

\frac{20}{3}

Decimal

6.66666 \dots

Solution steps

\int_{4}^{9} \frac{t - 3}{t} d t

Expand

\frac{t - 3}{t} : t - \frac{3}{t}

= \int_{4}^{9} t - \frac{3}{t} d t

Apply the Sum Rule:

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

= \int_{4}^{9} t d t - \int_{4}^{9} \frac{3}{t} d t

\int_{4}^{9} t d t = \frac{38}{3}

\int_{4}^{9} \frac{3}{t} d t = 6

= \frac{38}{3} - 6

Simplify

\frac{38}{3} - 6 : \frac{20}{3}

= \frac{20}{3}

x \to 0 lim (cos^{2} (5 x))

x \to - \infty lim (6)

\int (\frac{1}{x}) d x

\frac{d}{d x} (2^{x} + x^{2})

\frac{\partial}{\partial x} (4 x^{2} + 9 y^{2} - 45)

What is the integral from 4 to 9 of (t-3)/(sqrt(t)) ?
The integral from 4 to 9 of (t-3)/(sqrt(t)) is 20/3