Area under a curve. Hello friends, today I’ll talk about the area under a curve. Have a look!!

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*How to get the area under a parametric curve*

Area under a curve_compressed

### The area under a curve

Suppose is the equation of any curve. Then the area bounded by the curve, the -axis and the ordinates and will be

Now I’ll solve some examples of that.

**Solved examples of the area under a curve **

Note: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

**Example 1**

According to Stroud and Booth (2013)*, “Find the area enclosed by the curve and , between and .”

**Solution**

Now here I have to find out the area enclosed by the curve and , between and .

Let’s say is the area enclosed by the curve between and .

And is the area enclosed by the curve between and .

**Step 1**

Thus according to the formula for the area enclosed by a curve,

Then I’ll use the standard * formulas in integration* to get

Next, I’ll substitute the limits. So that gives

Now I can rewrite as

Also the value of is . So the value of becomes

Next, I’ll simplify it to get

Now I’ll get the value of .

**Step 2**

So the value of is

Next, I’ll integrate it to get

Then I’ll substitute the limits to get

Next, I’ll simplify it. So that gives

Therefore the area between the two curves is

Hence I can conclude that this is the answer to the given example.

Now I’ll give you another example.

**Example 2**

According to Stroud and Booth (2013)*, “Find the area bounded by the curve , the – axis and the ordinates at and .”

**Solution**

Now here the given curve is

First of all, I’ll rewrite it in terms of . So that means

Thus the area bounded by this curve, the – axis and the ordinates at and is

Now in order to integrate this function, I’ll use the method of integration by partial fraction. So first of all, I’ll get the partial fraction of .

**Step 1**

As I can see, here the function has equal degrees both in the numerator and the denominator. So I’ll rewrite as

Since , I can say

Now in order to get the partial fraction of , I’ll get the partial fraction of . Since the numerator has lower degree than that of the denominator, I’ll use the method of the * partial fraction of expressions with lower degree numerator*.

So that gives

where and are constants. Now my job is to get the values of and .

Hence I’ll simplify the right-hand side of this equation to get

Next, I’ll separate the coefficients to get

Then I’ll compare the coefficients of and the constants on both sides. And that gives me two different equations as

(1)

(2)

Now I’ll add equations (1) and (2) to get

Then I’ll put back in equation (2) to get

Therefore I can say that

Thus the value of is

Next, I’ll integrate to get the area bounded by this curve, the – axis and the ordinates at and .

**Step 2**

So this gives

And this means

Now that gives

Next, I’ll substitute the limits to get

Then I’ll simplify it to get

Since , I can say that

Also, we all know that . Thus the required area is

Now I’ll simplify it to get

Hence I can conclude that this is the answer to the given example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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