Hello Angelina Lopez,
1.) The Midpoint Rule is the approximation $$\int_a^b f(x)\,dx\approx M_n$$ where:
$$M_n=\frac{b-a}{n}\left(f\left(\frac{x_0+x_1}{2} \right)+f\left(\frac{x_1+x_2}{2} \right)+\cdots+f\left(\frac{x_{n-1}+x_n}{2} \right) \right)$$
In our case, we have:
$$a=1,\,b=3,\,n=4,\,f(x)=\frac{1}{x},\,x_k=a+\frac{b-a}{n}k=\frac{k+2}{2}$$
Hence:
$$\frac{x_{k-1}+x_k}{2}=\frac{\dfrac{(k-1)+2}{2}+\dfrac{k+2}{2}}{2}=\frac{2k+3}{4}$$
and so:
$$f\left(\frac{x_{k-1}+x_k}{2} \right)=\frac{4}{2k+3}$$
Thus:
$$M_4=\frac{3-1}{4}\sum_{k=1}^4\left(\frac{4}{2k+3} \right)=2\sum_{k=1}^4\left(\frac{1}{2k+3} \right)$$
$$M_4=2\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+ \frac{1}{11} \right)=\frac{3776}{3465}$$
2.) We are given:
(1) $$f(-x)=f(x)$$
(2) $$\int_0^6 f(x)\,dx=96$$
(3) $$\int_0^{-2} f(x)\,dx=10$$
and we are asked to find:
$$\int_{-2}^{-6} f(x)\,dx$$
From (1) and (2) we know:
$$\int_{-6}^{0} f(x)\,dx=96$$
which we may write as:
$$\int_{-6}^{-2} f(x)\,dx+\int_{-2}^{0} f(x)\,dx=96$$
Using $$\int_a^b g(x)\,dx=-\int_b^a g(x)\,dx$$ this becomes:
$$-\int_{-2}^{-6} f(x)\,dx-\int_{0}^{-2} f(x)\,dx=96$$
Using (3), we obtain:
$$-\int_{-6}^{-2} f(x)\,dx-10=96$$
Hence:
$$\int_{-6}^{-2} f(x)\,dx=-106$$
