# Calculating Area Under a Curve Using Change of Variables and Quotient Rule

• chwala
In summary, we discussed the approach of utilizing change of variables and quotient rule to find the area subtended by the normal, x-axis, and y-axis in a given problem. The solution involved integrating by letting u = sin 2x and using the fact that the gradient of the curve at x = π/4 is -4. The final result was ln(9/4) + π²/128.
chwala
Gold Member
Homework Statement
see attached question and ms
Relevant Equations
integration
Find question here;

Find solution here;

I used the same approach as ms- The key points to me were;

* making use of change of variables...

$$A_{1}=\int_0^\frac{π}{4} {\frac {4\cos 2x}{3-\sin 2x}} dx=-2\int_3^{-2} {\frac {du}{u}}= 2\int_2^3 {\frac {du}{u}}=2\ln 3-2\ln2=\ln 9 - \ln 4=\ln \frac{9}{4}$$

Now to find the other area subtended by the normal, x- axis and y- axis;
We make use of quotient rule. $$y'=\frac{(3-\sin 2x)(-8\sin 2x)-(4\cos 2x)(-2\cos 2x)}{(3-\sin 2x)^2}$$

the x-value =##\frac{π}{4}## can be substituted here directly and this is what i wanted to share. The gradient of the curve at the point where ##x=\frac{π}{4}## will be
$$y'=\frac{(3-1)(-8)-(0)(0)}{(3-1)^2}=\frac{2⋅-8}{2^2}=-4$$
Now for the given straight line equation, which is a Normal to the curve; We know that,
##y=mx +c##. It follows that,
##0=\frac{1}{4}⋅\frac{π}{4}+c##
##⇒ c=-\frac{π}{16}##
$$A_{2}=\frac {1}{2}⋅\frac{1}{4}⋅ \frac{π}{16}=\frac{π^2}{128}$$
$$A_{Required}= A_{1}+A_{2}=\ln \frac{9}{4}+ \frac{π^2}{128}$$

Last edited:
Apologies, is there a question here? Your answer agrees with the solution - so it looks fine to me...

Delta2
Master1022 said:
Apologies, is there a question here? Your answer agrees with the solution - so it looks fine to me...
Sometimes what happens in math solutions, is that you do one mistake, followed by a second mistake and you arrive at a correct result. However I can't find any mistakes in his solution and since it agrees with the official solution, it probably is correct.

Master1022
No question rather just sharing ...as highlighted in blue...maybe I should ask if there is another approach ...particularly on the integrating part of the problem.

Last edited:
As easy as ##\frac{\pi}{4}##

chwala
We could also integrate by;
Letting ##u=\sin 2x##
$$A_{1}=\int_0^\frac{π}{4} {\frac {4\cos 2x}{3-\sin 2x}} dx=2\int_0^1 {\frac {du}{3-u}}=-2\ln(3-1)+2\ln(3-0)=2\ln 3-2\ln2 ...$$

neilparker62

## 1. What is the purpose of finding the area under a curve?

The area under a curve is a measure of the total amount of a quantity represented by the curve. It is often used in mathematics and science to calculate the total value of a function or to determine the magnitude of a physical phenomenon.

## 2. How is the area under a curve calculated?

The area under a curve can be calculated using various methods, such as numerical integration, geometric formulas, or using mathematical software. The most common method is to divide the curve into small rectangles, calculate the area of each rectangle, and then add them together to get an approximation of the total area.

## 3. What is the difference between finding the area under a curve and finding the area enclosed by a curve?

Finding the area under a curve refers to calculating the total area between the curve and the x-axis, while finding the area enclosed by a curve refers to calculating the area inside a closed curve. The latter may involve subtracting the area under the curve from the total area enclosed by the curve.

## 4. Can the area under a curve be negative?

Yes, the area under a curve can be negative if the curve dips below the x-axis. This indicates that the quantity represented by the curve has a negative value. In such cases, the area under the curve is calculated by subtracting the negative area from the positive area.

## 5. What real-life applications use the concept of finding the area under a curve?

The concept of finding the area under a curve has many real-life applications, such as calculating the total displacement of an object over time, determining the total amount of a drug in a person's bloodstream, or estimating the total volume of a liquid in a container based on its changing depth over time. It is also used in fields such as economics, physics, and engineering to analyze and predict various phenomena.

Replies
2
Views
982
Replies
6
Views
1K
Replies
7
Views
1K
Replies
16
Views
2K
Replies
9
Views
496
Replies
6
Views
1K
Replies
3
Views
1K
Replies
1
Views
645
Replies
3
Views
1K
Replies
12
Views
1K