Question

A graph of $f(x) = \frac{1}{(8 - x)^{0.9}}$ follows. The improper integral $\int_0^8 \frac{1}{(8 - x)^{0.9}} dx$ \textcircled{diverges} \textcircled{converges}. If it is convergent, then its area = \boxed{ }

          A graph of $f(x) = \frac{1}{(8 - x)^{0.9}}$ follows.
The improper integral $\int_0^8 \frac{1}{(8 - x)^{0.9}} dx$ \textcircled{diverges} \textcircled{converges}.
If it is convergent, then its area = \boxed{ }

A graph of f(x) = (1)/((8 - x)^0.9) follows.
The improper integral ∫0^8 (1)/((8 - x)^0.9) dx diverges converges.
If it is convergent, then its area =

Added by Kevin S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Eric Mockensturm

10/03/2021

Step 1

Then $$ I = \int_0^8 \frac{1}{(8-x)^{0.9}} dx $$ Let $u = 8 - x$. Then $du = -dx$, so $dx = -du$. When $x = 0$, $u = 8$. When $x = 8$, $u = 0$. Therefore, $$ I = \int_8^0 \frac{1}{u^{0.9}} (-du) = \int_0^8 u^{-0.9} du $$ $$ I = \left[ \frac{u^{1-0.9}}{1-0.9} Show more…

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A graph of $f(x) = \frac{1}{(8-x)^{0.9}}$ follows. The improper integral $\int_0^8 \frac{1}{(8-x)^{0.9}} dx$ $\bigcirc$ diverges $\bigcirc$ converges. If it is convergent, then its area = $\boxed{}$