What is the integral from 3 to 4 of-log_{9}(x) ?

The integral from 3 to 4 of-log_{9}(x) is -8log_{9}(2)+3/2+1/(2ln(3))

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integral from 3 to 4 of-log_{9}(x)

\int_{3}^{4} - lo g_{9} (x) d x

- 8 lo g_{9} (2) + \frac{3}{2} + \frac{1}{2 ln ( 3 )}

Decimal

- 0.56859 \dots

Solution steps

\int_{3}^{4} - lo g_{9} (x) d x

Take the constant out:

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

= - \int_{3}^{4} lo g_{9} (x) d x

Apply Integration By Parts

= - [x lo g_{9} (x) - \int \frac{1}{2 ln ( 3 )} d x]_{3}^{4}

\int \frac{1}{2 ln ( 3 )} d x = \frac{1}{2 ln ( 3 )} x

= - [x lo g_{9} (x) - \frac{1}{2 ln ( 3 )} x]_{3}^{4}

Compute the boundaries:

8 lo g_{9} (2) - \frac{3}{2} - \frac{1}{2 ln ( 3 )}

= - (8 lo g_{9} (2) - \frac{3}{2} - \frac{1}{2 ln ( 3 )})

Simplify

= - 8 lo g_{9} (2) + \frac{3}{2} + \frac{1}{2 ln ( 3 )}

\int 2 x^{2} + \frac{1}{( 2 x ) ^{2}} d x

y \to + 0 lim (\frac{y}{sin ( 5 y )})

d er i v a t i v e g (x) = x^{\frac{2}{3}}

\int_{0}^{3} t e^{- 1} d t

t a y l or e^{- x^{2}}

What is the integral from 3 to 4 of-log_{9}(x) ?
The integral from 3 to 4 of-log_{9}(x) is -8log_{9}(2)+3/2+1/(2ln(3))