What is the integral from 7 to 8 of 6log_{9}(t) ?

The integral from 7 to 8 of 6log_{9}(t) is 6(24log_{9}(2)-7log_{9}(7)-1/(2ln(3)))

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integral from 7 to 8 of 6log_{9}(t)

\int_{7}^{8} 6 lo g_{9} (t) d t

6 (24 lo g_{9} (2) - 7 lo g_{9} (7) - \frac{1}{2 ln ( 3 )})

Decimal Notation

5.50010 \dots

Solution steps

\int_{7}^{8} 6 lo g_{9} (t) d t

Take the constant out:

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

= 6 \cdot \int_{7}^{8} lo g_{9} (t) d t

Apply Integration By Parts

: [t lo g_{9} (t) - \int \frac{1}{2 ln ( 3 )} d t]_{7}^{8}

= 6 [t lo g_{9} (t) - \int \frac{1}{2 ln ( 3 )} d t]_{7}^{8}

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

= 6 [t lo g_{9} (t) - \frac{1}{2 ln ( 3 )} t]_{7}^{8}

Compute the boundaries:

24 lo g_{9} (2) - 7 lo g_{9} (7) - \frac{1}{2 ln ( 3 )}

= 6 (24 lo g_{9} (2) - 7 lo g_{9} (7) - \frac{1}{2 ln ( 3 )})

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What is the integral from 7 to 8 of 6log_{9}(t) ?
The integral from 7 to 8 of 6log_{9}(t) is 6(24log_{9}(2)-7log_{9}(7)-1/(2ln(3)))