Area Between Two Curves: The most general class of problems in calculating the area of bounded regions is the one involving the regions between two curves. All other problems can be treated as its subset. Besides being an important application of definite integrals, these problems also gives one an idea about the nature of the two curves involved.

**Browse more Topics under Application Of Integrals**

If plotting one of the curves is notoriously difficult, one can get the idea of the nature of the curve by calculating the area of the region bounded by the curve with some other known curve! Let us now look at the method we will employ to solve the area between two curves problems!

### Suggested Videos

## General Formula for Area Between Two Curves

If we have two given curves:

- P: y = f(x)
- Q: y = g(x)

The first and the most important step is to plot the two curves on the same graph. If one can’t plot the exact curve, at least an idea of the relative orientations of the curves should be known. Anyway, let the graph look something like this:

You should be able to see that one needs to calculate the coordinates of the points of intersection of the two curves to find out the boundaries of the region whose area has to be calculated. Let the bounding values of x, in this case, be x_{1} and x_{2}. Then the formula for the area A of the region bounded between the two curves is:

$$ {A = \int_{x_1}^{x_2} [f(x) – g(x)]dx} $$

–> Similarly, if the form of the equations of the curves is such that we have:

- P: x = f(y)
- Q: x = g(y)

The formula for the area A of the region bounded between the two curves within the domain y = y_{1} and y_{2}, then reduces to the form: $$ {A = \int_{y_1}^{y_2} [f(y) – g(y)]dy} $$. There are various important things to keep in mind when applying these formulae directly. I’m highlighting them below.

## General Conventions on Area Between Two Curves

When applying the formula for the area between two curves \({A = \int_{x_1}^{x_2} [f(x) – g(x)]dx} \)

- The upper function in the graph i.e. the one with the greater value of y for a given x is taken to be f(x)
- The lower function in the graph i.e. the one with a smaller value of y for a given x is taken to be g(x)
- The upper and the lower functions may be different for two different regions on the graph. It is important to calculate the area for such regions separately.
- The area above the y-axis is allotted a positive sign.
- The area below the y-axis is allotted a negative sign.

Similarly, we have certain conventions to be followed when we use the formula \( {A = \int_{y_1}^{y_2} [f(y) – g(y)]dy} \)

- The right function in the graph i.e. the one with the greater value of x for a given y is taken to be f(y)
- The left function in the graph i.e. the one with a smaller value of x for a given y is taken to be g(y)
- The right and the left functions may be different for different regions on the graph. It is important to calculate the area for such regions separately.
- The area on the right side of the x-axis is allotted a positive sign.
- The area on the left side of the x-axis is allotted a negative sign.

Now let’s solve some questions to learn how to use the general formula with the necessary conventions!

## Solved Examples for You

**Question 1:** Calculate the total area of the region bounded between the curves y = 6x – x^{2} and y = x^{2}.

**Answer :** The intersection points of the curve can be solved by putting the value of y = x^{2} into the other equation. We then get:

x^{2} = 6x – x^{2}

2x^{2} -6x = 0

2x(x – 3) = 0

x = 0 / x = 3

Correspondingly we can get the values of y as 0/9. Thus the intersection points can be given as P(3,9) and Q(0,0). The system is as shown below-

From the graph, the upper curve will be f(x) and the lower one will be g(x). So we have f(x) = 6x – x^{2 }and g(x) = x^{2}. Then the area:

$$ {A = \int_{x_1}^{x_2} [f(x) – g(x)]dx} $$

$$ {A = \int_{0}^{3} [6x – x^2 – x^2]dx} $$

$$ {A = \int_{0}^{3} [6x – 2x^2]dx} $$

$$ {A = [6\times{\frac{x^2}{2}} – 2\times{\frac{x^3}{3}}]_{0}^{3}} $$

$$ {A = (6 \times{\frac{9}{2}} + 2 \times{\frac{27}{3}}) – (6 \times{\frac{0}{2}} + 2 \times{\frac{0}{3}}) }$$

$$ {A = (27 + 18) – (0 + 0)}$$

$$ {A = 45}$$

Since the area is above the y-axis, we will take it with a positive sign. Thus the answer is A = 45.

### Question 3

Calculate the area of the region bounded by the curves y^{2} + 4x – y = 0 and the straight line y = x.

**Answer :** Note here that it is very complex to express the equation y^{2} + 4x – y = 0 in the form of y = f(x); however, the form x = f(y) can easily be obtained. $$ {x = \frac{- y^2 + y}{4}} $$ Thus here we must use the alternative formula for the calculation for the area under the curve. Now let us calculate the intersection points first. Using the value y = x in the equation of the curve, we get

y^{2} + 4y – y = 0

y^{2} + 3y = 0

y(y + 3) = 0

y = 0/ -3

Corresponding to the values of y, one can get x = 0/ -3. Thus the points of intersection are P(-3,-3) and Q(0,0). The graph for the system can be drawn as:

From the graph, the curve on the right will be f(y) and the curve on the left will be g(y). So we have: \({f(y) = \frac{- y^2 + y}{4}}\)^{ }and g(y) = y. ^{ }Then the area:

$$ {A = \int_{x_1}^{x_2} [f(y) – g(y)]dy} $$

$$ {A = \int_{0}^{3} [\frac{- y^2 + y}{4} – y]dy} $$

$$ {A = \int_{0}^{3} [-\frac{y^2}{4} – \frac{3}{4}y]dy} $$

$$ {A = [-\frac{y^3}{4 \times {3}} – \frac{3}{4} \times {\frac{y^2}{2}}]_{0}^{3}} $$

$$ {A = -[(0 + 0) – (\frac{-27}{12} + \frac{3}{4} \times{\frac{9}{2}})] } $$

$$ {A = \frac{27}{24}} $$

$$ \text{Area as specified by the definite integral =} {- \frac{27}{24}} $$ where in the last line we have employed the convention, if the region lies to the left of the y-axis, the area wll be taken with a negative sign.

**Question 3: Is it possible for the area between two curves to be negative?**

**Answer:** No, the area between the two curves will not be negative. This is because the area between the two curves is not like the area under the curve. Hence, area between the two curves shall always be positive.

**Question 4: Is it possible for an integral to be negative?**

**Answer: **Yes, it is possible for a definite integral to be negative. In case all the area within the interval is below the x-axis but still above the curve, then the negative result will take place.

**Question 5: Explain the area under the curve?**

**Answer:** The area under a curve that exists between two points can be calculated by conducting a definite integral between the two points. To calculate the area under the curve y = f(x) between x = a & x = b, one must integrate y = f(x) between the limits of a and b.

**Question 6: What is meant by the polar curve?**

**Answer:** A polar curve refers to a shape whose construction takes place by using the polar coordinate system. They are characterized by points that exist at a variable distance from the pole or origin.

## Leave a Reply