Problem 11. You are to construct an open rectangular box with a square base and a volume of 48 cm³. If material for the bottom costs 6/cm² and material for the sides costs 4/cm², what dimensions will result in the least expensive box? What is the minimum cost? Support your answers with diagrams.
Problem 12. In a maximum power transfer theorem scenario, the power delivered to a load $P(R_L)$ is expressed as
$P(R_L) = \frac{V^2 R_L}{(R+R_L)^2}$,
where $V$ is the source voltage and $R$ is the internal resistance. Determine the load resistance $R_L$ that maximizes the power delivered to the load.
Problem 13. Use lower and lower sums with four rectangles to estimate the area under the graph of the functions.
(a) $f(x) = x^2$ between $x = 0$ and $x = 1$
(b) $f(x) = x^3$ between $x = 0$ and $x = 1$
(c) $f(x) = 1/x$ between $x = 1$ and $x = 5$
(d) $f(x) = 4 - x^2$ between $x = -2$ and $x = 2$
Problem 14. Use known area formulas to evaluate the integrals
(a) $\int_{-3}^{3} \sqrt{9 - x^2} dx$
(b) $\int_{-2}^{2} |x| dx$
(c) $\int_{0}^{b} 4x dx$, $b > 0$
(d) $\int_{-1}^{1} (1 + \sqrt{1 - x^2}) dx$
Problem 15. Find the derivative $dy/dx$ if
(a) $y = \int_{0}^{x} \cos t dt$
(b) $y = \int_{0}^{x^3} t^{-2/3} dt$
(c) $y = \int_{0}^{\tan x} \sec^2 t dt$
(d) $y = \int_{1}^{x} \sqrt{1 + t^2} dt$
(e) $y = \int_{0}^{x^2} \sin(t^2) dt$
(f) $y = \int_{x^2}^{x} \sin(t^3) dt$
(g) $y = \int_{0}^{\sin x} \frac{dt}{\sqrt{1 - t^2}}$, $|x| < \frac{\pi}{2}$
Problem 16. Find $f(4)$ if $\int_{0}^{x} f(t) dt = x \cos \pi x$.
Problem 17. Use integration by substitution to evaluate the integrals.
(a) $\int_{-2}^{2} \frac{x^2}{(x^2 + 3)^4} dx$
(b) $\int_{1}^{4} (x^2 + \sqrt{x}) dx$
(c) $\int_{0}^{\pi/2} \frac{1}{1 + \cos 2x} dx$
(d) $\int_{\pi/4}^{\pi/2} (4 \sec^2 x + \frac{\pi}{x}) dx$
(e) $\int_{0}^{\sqrt{\pi/2}} x \cos x^2 dx$
(f) $\int_{0}^{\pi/2} \frac{\sin x}{\sqrt{2 - \cos^2 x}} dx$
(g) $\int_{1}^{5} \frac{x dx}{\sqrt{x^2 + 4}}$
(h) $\int_{-4}^{4} |x| dx$
(i) $\int_{0}^{3} \sqrt{3 - 2x} dx$
(j) $\int_{0}^{1} \sec^2 (3x + 2) dx$
(k) $\int_{0}^{\pi/2} \cos(\frac{\pi}{2} - x) dx$
(l) $\int_{1}^{2} \sqrt{2x - \frac{1}{x}} dx$
