5. The graphs of the functions $f$ and $g$ are shown in the figure for $0 \leq x \leq 3$. It is known that $g(x)=\frac{12}{3+x}$ for $x \geq 0$. The twice-differentiable function $f$, which is not explicitly given, satisfies $f(3)=2$ and $\int_{0}^{3} f(x) d x=10$.
(a) Find the area of the shaded region enclosed by the graphs of $f$ and $g$.
(b) Evaluate the improper integral $\int_{0}^{\infty}(g(x))^{2} d x$, or show that the integral diverges.
(c) Let $h$ be the function defined by $h(x)=x \cdot f^{\prime}(x)$. Find the value of $\int_{0}^{3} h(x) d x$.

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Accepted Answer

∻Solution by Steps∻ ### Part (a) ‖ step 1 ⋮ To find the area of the shaded region between the graphs of $f(x)$ and $g(x)$ from $x = 0$ to $x = 3$, we need to compute the integral of the difference between $f(x)$ and $g(x)$. ‖ ‖ step 2 ⋮ The area $A$ is given by the integral: $A = \int_{0}^{3} [f(x) - g(x)] \, dx$. ‖ ‖ step 3 ⋮ We know $g(x) = \frac{12}{3+x}$ and $\int_{0}^{3} f(x) \, dx = 10$. ‖ ‖ step 4 ⋮ Compute $\int_{0}^{3} g(x) \, dx$: $$ \int_{0}^{3} \frac{12}{3+x} \, dx = 12 \int_{0}^{3} \frac{1}{3+x} \, dx = 12 \ln(3+x) \Big|_{0}^{3} = 12 \ln(6) - 12 \ln(3) = 12 \ln(2) $$ ‖ ‖ step 5 ⋮ The area $A$ is then: $$ A = \int_{0}^{3} f(x) \, dx - \int_{0}^{3} g(x) \, dx = 10 - 12 \ln(2) $$ ‖ ### Part (b) ‖ step 1 ⋮ To evaluate the improper integral $\int_{0}^{\infty} (g(x))^2 \, dx$, we first express $g(x)$: $$ g(x) = \frac{12}{3+x} $$ $$ (g(x))^2 = \left( \frac{12}{3+x} \right)^2 = \frac{144}{(3+x)^2} $$ ‖ ‖ step 2 ⋮ The integral becomes: $$ \int_{0}^{\infty} \frac{144}{(3+x)^2} \, dx $$ ‖ ‖ step 3 ⋮ Use the substitution $u = 3 + x$, hence $du = dx$: $$ \int_{3}^{\infty} \frac{144}{u^2} \, du = 144 \int_{3}^{\infty} u^{-2} \, du = 144 \left[ -u^{-1} \right]_{3}^{\infty} = 144 \left( 0 - (-\frac{1}{3}) \right) = 48 $$ ‖ ### Part (c) ‖ step 1 ⋮ To find $\int_{0}^{3} h(x) \, dx$ where $h(x) = x \cdot f'(x)$, we use integration by parts. Let $u = x$ and $dv = f'(x) \, dx$. ‖ ‖ step 2 ⋮ Then $du = dx$ and $v = f(x)$. Using integration by parts: $$ \int_{0}^{3} x f'(x) \, dx = \left[ x f(x) \right]_{0}^{3} - \int_{0}^{3} f(x) \, dx $$ ‖ ‖ step 3 ⋮ Evaluate the boundary term: $$ \left[ x f(x) \right]_{0}^{3} = 3 f(3) - 0 \cdot f(0) = 3 \cdot 2 = 6 $$ ‖ ‖ step 4 ⋮ We know $\int_{0}^{3} f(x) \, dx = 10$, so: $$ \int_{0}^{3} x f'(x) \, dx = 6 - 10 = -4 $$ ‖ ∻Answer∻ ⚹ (a) $10 - 12 \ln(2)$ ⚹ ⚹ (b) $48$ ⚹ ⚹ (c) $-4$ ⚹ ∻Key Concept∻ ⚹Area between curves, improper integrals, and integration by parts⚹ ∻Explanation∻ ⚹The area between curves is found by integrating the difference of the functions. Improper integrals require limits to evaluate. Integration by parts is used to integrate products of functions.⚹◊What is the expression for $g(x)$ in the given scenario involving the functions $f$ and $g$?⍭ Generate me a similar question◊

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