# Angelina Lopez's questions at Yahoo Answers regarding definite integrals

• MHB
• MarkFL
In summary, we discussed using the Midpoint Rule to approximate the integral of a function on a given interval, and also used the properties of even functions to find the integral on a different interval.
MarkFL
Gold Member
MHB
Here are the questions:

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.

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:

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

## 1. What is a definite integral?

A definite integral is a mathematical concept that represents the area under a curve between two points on a graph. It is denoted by ∫a^b f(x) dx, where 'a' and 'b' are the lower and upper limits of integration and 'f(x)' is the function being integrated.

## 2. How is a definite integral different from an indefinite integral?

A definite integral has specific limits of integration and gives a numerical value, while an indefinite integral does not have limits and gives a function as a result. Essentially, a definite integral is a real number, while an indefinite integral is a function.

## 3. What is the purpose of finding definite integrals?

The main purpose of finding definite integrals is to calculate the area under a curve in a given interval. It is also used to solve various real-life problems in fields such as physics, engineering, and economics.

## 4. How do you solve a definite integral?

To solve a definite integral, you can use various integration techniques such as substitution, integration by parts, and trigonometric substitution. Once you have found the antiderivative, you can plug in the upper and lower limits of integration and evaluate the expression to get the numerical value.

## 5. Can definite integrals be negative?

Yes, definite integrals can be negative. This occurs when the function being integrated has a negative value in the given interval. The negative sign in front of the integral denotes that the area under the curve is below the x-axis, resulting in a negative value.

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