Question 1. Graphical Integration Challenge: u-Substitution and Integration by Parts Let $F(x)=\int_{-1}^{x} f(t) \mathrm{d} t$ and $G(x)=\int_{0}^{x} g(t) \mathrm{d} t$, where the graphs of $f(x)$ and $g(x)$ on $[-2,4]$ are below. (a) Evaluate $\int_{0}^{2}[F(x) f(x)-G(x) g(x)] \mathrm{d} x$. HINT: Consider a $u$-substitution (possibly multiple substitutions). You may need to invoke Part I of the Fundamental Theorem of Calculus.
Solution
Step 1 ：Let $F(x)=\int_{-1}^{x} f(t) \mathrm{d} t$ and $G(x)=\int_{0}^{x} g(t) \mathrm{d} t$, where the graphs of $f(x)$ and $g(x)$ on $[-2,4]$ are given.
Step 2 ：We are asked to evaluate $\int_{0}^{2}[F(x) f(x)-G(x) g(x)] \mathrm{d} x$.
Step 3 ：We can use the Fundamental Theorem of Calculus and u-substitution to simplify the integral. The Fundamental Theorem of Calculus states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then $\int_{a}^{b} f(x) dx = F(b) - F(a)$.
Step 4 ：For the first term, let $u = F(x)$ and $dv = f(x)dx$. Then find $du$ and $v$, and use the formula for integration by parts, $\int u dv = uv - \int v du$, to simplify the integral.
Step 5 ：For the second term, let $u = G(x)$ and $dv = g(x)dx$. Then find $du$ and $v$, and use the formula for integration by parts, $\int u dv = uv - \int v du$, to simplify the integral.
Step 6 ：However, without the actual graphs of $f(x)$ and $g(x)$, we cannot provide a numerical answer.
Step 7 ：\(\boxed{\text{The final answer would depend on the specific graphs of } f(x) \text{ and } g(x).}\)
