# Integration of a function of x

• chwala
In summary, the conversation is about solving the integral of 45/3x^2(4-2x)^3dx and whether to use integration by parts or substitution. One user suggests using integration by parts three times, while another suggests expanding and multiplying out the factors or using substitution. The latter user shows their work for using substitution and concludes that it will give the correct solution.
chwala
Gold Member

## Homework Statement

##∫45.1/3x^2 (4-2x)^3dx##[/B]

## The Attempt at a Solution

##45/3∫x^2(4-2x)^3dx = let u = x^2 du= 2x, dv= (4-2x)^3 v=(2-x)/-4 ##

using intergration by parts is this right[/B]

It's not obvious whether you mean ##\frac{45.1}{3x^2 (4 - 2x) ^3}## or ##\frac{45.1}{3}x^2 (4-2x)^3##.

If you want to integrate ##\int \frac{45}{3} x^2(4-2x)^3dx=15 \int x^2(4-2x)^3dx## by parts, you will have to apply it three times and in the end, the polynomial has to be of degree ##6##, for you started with a degree ##5## polynomial. I'm rather sure that performing the multiplication ##x^2(4-2x)^3## and integrating term by term is almost faster.

The formula by parts goes ##\int u'v = uv - \int uv'##

chwala said:

## Homework Statement

##∫45.1/3x^2 (4-2x)^3dx##[/B]

## The Attempt at a Solution

##45/3∫x^2(4-2x)^3dx = let u = x^2 du= 2x, dv= (4-2x)^3 v=(2-x)/-4 ##

using intergration by parts is this right[/B]
If you are solving ##\int x^2 (4-2x)^3 dx## then integration by parts is perhaps overkill. You could simply expand out all terms or first make the substitution ##u = 4-2x## which means you no longer have to deal with expanding out a cubic power. If you have to use integration by parts, your integration of v is incorrect.

hilbert2 said:
It's not obvious whether you mean ##\frac{45.1}{3x^2 (4 - 2x) ^3}## or ##\frac{45.1}{3}x^2 (4-2x)^3##.
sorry i mean the latter...

I expanded and multiplied out the factors...and got it as follows,

##45/3∫x^2(16-16x+4x^2)(4-2x) = 45/3∫x^2(64-32x-64x+32x^2+16x^2-8x^3)dx = 45/3∫(-8x^5+48x^4-96x^3+64x^2)dx##

this is an easy integral i was just tired, i will try use the substitution method and see what comes out.
now on using substitution
let ##u=4-2x, →dx=du/-2, x= (4-u)/2 ⇒x^2=(16-8u+u^2)/4,

⇒45/3∫((u^2-8u+16)/4)). (u^3) (du/-2) = 45/-24∫(u^5-8u^4+16u^3)du ##

which will give correct solution as the other long method.
Thank you guys, greetings from Africa.

Last edited:
@chwala, please post questions on calculus in the Calculus & Beyond section, not the Precalculus section, where you originally posted this thread.

Mark44 said:
@chwala, please post questions on calculus in the Calculus & Beyond section, not the Precalculus section, where you originally posted this thread.
OK sir

## 1. What is the definition of integration?

Integration is a mathematical process that involves finding the area under a curve on a graph. It is the inverse operation of differentiation, which is finding the slope of a curve at a given point.

## 2. How do you perform integration?

To perform integration, you can use various techniques such as the substitution method, integration by parts, or using specific integration rules for common functions. The most common method is using the integral symbol (∫) and solving the integral using basic algebraic operations.

## 3. What are the applications of integration?

Integration has many practical applications in fields such as physics, engineering, economics, and statistics. It is used to calculate the area under a curve, compute volumes and areas of irregular shapes, and solve problems involving rates of change.

## 4. What is the difference between definite and indefinite integration?

Definite integration involves finding the numeric value of the area under a curve between two specific points on a graph. On the other hand, indefinite integration results in a function that represents the area under a curve without specifying any limits. It is also known as the antiderivative of a function.

## 5. How is integration related to differentiation?

Integration and differentiation are inverse operations of each other. This means that if a function is differentiated, the resulting function can be integrated back to the original function. Integration can also be used to check the accuracy of a differentiation calculation.

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