## Calculus: Early Transcendentals 8th Edition

$$\frac{45}{4} - \ln8$$
Area of a graph is = $\int^b_a[f(x) - g(x)] dx$ where f(x) is the graph above g(x). In this question, $f(x) = \sqrt[3] x$ and $g(x) = \frac{1}{x}$ For our limits of integration $a$ = the x coord of the left intercept (where f(x) = g(x)) and $b$ = the line $x=8$ $$\int^b_a f(x)-g(x) dx$$ $$\int^8_1 \sqrt[3] x - \frac{1}{x} dx$$ Now we must solve the definite integral: $$= \int^8_1 x^{\frac{1}{3}} - \frac{1}{x} dx$$ $$= (\frac{3}{4}x^{\frac{4}{3}} - \ln|x| dx)|^8_1$$ Plug in the limits of integration and simplify (note that $\ln1 = 0$): $$= [\frac{3}{4}(8)^{\frac{4}{3}} - \ln 8] - [\frac{3}{4}(1)^{\frac{4}{3}} - \ln 1]$$ $$= 12 - \ln8 - \frac{3}{4}$$ $$= \frac{45}{4} - \ln8$$