Angelina Lopez's questions at Yahoo Answers regarding definite integrals

[MarkFL]

Here are the questions:

Calculus integral problems. Please help I would appreciate.?


1. Consider f(x)= 1/x on [1,3]. Use four sub-intervals to approximate the integral from 1 to 3 f(x)dx by using the Midpoint Rule.

2. Suppose f(x) is a continuous even function and that the integral from 0 to 6 f(x)dx=96 and the integral from 0 to -2 f(x)dx =10. Find the integral from -2 to -6 f(x)dx.

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

Hello Angelina Lopez,

1.) The Midpoint Rule is the approximation \(\displaystyle \int_a^b f(x)\,dx\approx M_n\) where:

\(\displaystyle M_n=\frac{b-a}{n}\left(f\left(\frac{x_0+x_1}{2} \right)+f\left(\frac{x_1+x_2}{2} \right)+\cdots+f\left(\frac{x_{n-1}+x_n}{2} \right) \right)\)

In our case, we have:

\(\displaystyle a=1,\,b=3,\,n=4,\,f(x)=\frac{1}{x},\,x_k=a+\frac{b-a}{n}k=\frac{k+2}{2}\)

Hence:

\(\displaystyle \frac{x_{k-1}+x_k}{2}=\frac{\dfrac{(k-1)+2}{2}+\dfrac{k+2}{2}}{2}=\frac{2k+3}{4}\)

and so:

\(\displaystyle f\left(\frac{x_{k-1}+x_k}{2} \right)=\frac{4}{2k+3}\)

Thus:

\(\displaystyle M_4=\frac{3-1}{4}\sum_{k=1}^4\left(\frac{4}{2k+3} \right)=2\sum_{k=1}^4\left(\frac{1}{2k+3} \right)\)

\(\displaystyle M_4=2\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+ \frac{1}{11} \right)=\frac{3776}{3465}\)

2.) We are given:

(1) \(\displaystyle f(-x)=f(x)\)

(2) \(\displaystyle \int_0^6 f(x)\,dx=96\)

(3) \(\displaystyle \int_0^{-2} f(x)\,dx=10\)

and we are asked to find:

\(\displaystyle \int_{-2}^{-6} f(x)\,dx\)

From (1) and (2) we know:

\(\displaystyle \int_{-6}^{0} f(x)\,dx=96\)

which we may write as:

\(\displaystyle \int_{-6}^{-2} f(x)\,dx+\int_{-2}^{0} f(x)\,dx=96\)

Using \(\displaystyle \int_a^b g(x)\,dx=-\int_b^a g(x)\,dx\) this becomes:

\(\displaystyle -\int_{-2}^{-6} f(x)\,dx-\int_{0}^{-2} f(x)\,dx=96\)

Using (3), we obtain:

\(\displaystyle -\int_{-6}^{-2} f(x)\,dx-10=96\)

Hence:

\(\displaystyle \int_{-6}^{-2} f(x)\,dx=-106\)