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

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

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

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

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

Decimal Notation

- 0.56859 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

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