(Solved): 1: Theorem 1 (The Mean Value Theorem for Integrals) If f(x) is continuous on [a,b], then there exi ...

1: Theorem 1 (The Mean Value Theorem for Integrals) If $$f(x)$$ is continuous on $$[a,b]$$, then there exists a number $$c$$ in $$[a,b]$$ such that $$f(c)=f_{\text{avg }}=\frac{1}{b?a}{?}_a^{b}f(x)dx$$ that is, $${?}_a^{b}f(x)dx=f(c)(b?a)$$ The Mean Value Theorem for Integrals is a consequence of the Mean Value Theorem for derivatives and the Fundamental Theorem of Calculus. The geometric interpretation of the Mean Value Theorem for Integrals is that, for positive functions $$f$$, there is a number c such that the rectangle with base $$[\mathrm{a},\mathrm{b}]$$ and height $$f(c)$$ has the same area as the region under the graph of $$f(x)$$ from $$a$$ to $$b$$, as shown in Figure 1. Figure 1: An illustration of the Mean Value Theorem for the integrals. One can always chop off the top of a (two-dimensional) mountain at a certain height (namely, $$f_{\text{avg }}$$ ) and use it to fill in the valleys so that the mountain becomes completely flat. Question (5 marks): Find the value $$c$$ to satisfy the mean value theorem for the integral $${?}_0^{1}x(1?x)dx$$.