# Find the value of the definite integral

• chwala
In summary: Giving something to three decimal places is not giving the exact value. (Unless the decimal expansion terminates there - which it doesn't in this case)On substituting the given limits, we end up with,[1.2990−0+2.094395]−[0]=3.393 to 3 decimal places...
chwala
Gold Member
Homework Statement
see attached
Relevant Equations
integration
Find question here,

My approach, using cosine sum and product concept, we shall have;

##\cos (A+B)-\cos (A-B)=-2\sin A\sin B##

##⇒\cos D-\cos C=-2\sin\dfrac{C+D}{2} \sin\dfrac {C-D}{-2}##

##⇒-3[\cos(A+B)-\cos(A-B)]=6\sin A sinB##

We are given ##A=4θ## and ##B=2θ##, therefore,

##⇒-3[\cos 6θ-\cos 2θ]=6\sin 4θ\sin2θ##

##⇒3\cos 2θ-3\cos 6θ=6\sin 4θ\sin2θ##

I made use of chain rule in my working... i.e letting ##u=2θ## and ##u=6θ## ...

##\int_0^\frac{π}{3} [3\cos 2θ-3\cos 6θ+2] dθ=\dfrac{3}{2} \sin 2θ-\dfrac{1}{2} \sin 6θ +2θ## from ##[θ=0]## to ##[θ=\frac{π}{3}]##

On substituting the given limits, we end up with,

##[1.2990-0+2.094395]-[0]=3.393## to 3 decimal places...

Any other approach is welcome...cheers

Last edited:
Delta2
Giving something to three decimal places is not giving the exact value. (Unless the decimal expansion terminates there - which it doesn’t in this case)

chwala and Mark44
chwala said:
On substituting the given limits, we end up with,

[1.2990−0+2.094395]−[0]=3.393 to 3 decimal places...
As @Orodruin pointed out, your answer doesn't comply with the problem's request to "find the exact value".

For example, if ##\theta = \frac \pi 3##, then the value of ##\sin(2\theta)## is ##\sin(\frac{2\pi}3)##. As an exact value, this is ##\frac{\sqrt 3}2##; as an approximate value this is ##\approx 0.866##.

chwala
Another approach would be make the substitution ##u=2\theta## and observe,
$$sin(2u)sin(u)=2\cos(u)\sin^2(u)$$
A further substitution should immediately come to mind.

Delta2 and SammyS
Thanks...in that case i will need to leave the final answer in terms of ##π## i will look at that...

Aaaaargh this was easy...I will post the exact solution later in the day...when I get hold of my laptop...

chwala said:
Homework Statement:: see attached
Relevant Equations:: integration

Find question here,
View attachment 301958

My approach, using cosine sum and product concept, we shall have;

##\cos (A+B)-\cos (A-B)=-2\sin A\sin B##

##⇒\cos D-\cos C=-2\sin\dfrac{C+D}{2} \sin\dfrac {C-D}{-2}##

##⇒-3[\cos(A+B)-\cos(A-B)]=6\sin A sinB##

We are given ##A=4θ## and ##B=2θ##, therefore,

##⇒-3[\cos 6θ-\cos 2θ]=6\sin 4θ\sin2θ##

##⇒3\cos 2θ-3\cos 6θ=6\sin 4θ\sin2θ##

I made use of chain rule in my working... i.e letting ##u=2θ## and ##u=6θ## ...

##\int_0^\frac{π}{3} [3\cos 2θ-3\cos 6θ+2] dθ=\dfrac{3}{2} \sin 2θ-\dfrac{1}{2} \sin 6θ +2θ## from ##[θ=0]## to ##[θ=\frac{π}{3}]##

On substituting the given limits, we end up with,

##[1.2990-0+2.094395]-[0]=3.393## to 3 decimal places...

Any other approach is welcome...cheers
We shall have;

##...\dfrac{3}{2} \sin \dfrac{2π}{3}-\dfrac{1}{2} \sin 2π+\dfrac{2π}{3}##
##=\dfrac{3}{2}⋅\dfrac{\sqrt 3}{2}+\dfrac{2π}{3}##
##=\dfrac{3\sqrt 3}{4}+\dfrac{2π}{3}##

Delta2

## 1. What is a definite integral?

A definite integral is a mathematical concept used to find the area under a curve between two points on a graph. It is represented by the symbol ∫ and is used to calculate the total value of a function over a specific interval.

## 2. How do you find the value of a definite integral?

To find the value of a definite integral, you must first determine the limits of integration, which are the starting and ending points on the graph. Then, you need to evaluate the function at these points and subtract the result of the lower limit from the result of the upper limit.

## 3. What is the difference between a definite and indefinite integral?

The main difference between a definite and indefinite integral is that a definite integral has specific limits of integration, while an indefinite integral does not. A definite integral results in a numerical value, while an indefinite integral results in a function.

## 4. Can you use a calculator to find the value of a definite integral?

Yes, you can use a calculator or a computer program to find the value of a definite integral. However, it is important to understand the concept and steps of finding the value by hand before relying on technology.

## 5. What are some real-life applications of finding the value of a definite integral?

Definite integrals have many real-life applications, such as calculating the area under a curve in physics, determining the total distance traveled in calculus, and finding the total cost or profit in economics. They are also used in engineering, biology, and other fields to model and analyze various phenomena.

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