Integration of a function of x

[chwala]

Homework Statement

$∫45.1/3x^2 (4-2x)^3dx$[/B]

Homework Equations

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]

 

[hilbert2]

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

 

[fresh_42]

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 $\int u'v = uv - \int uv'$

 

[CAF123]

Homework Statement

$∫45.1/3x^2 (4-2x)^3dx$[/B]

Homework Equations

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.

 

[chwala]

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...

 

[chwala]

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.

 

[Mark44]

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

 

[chwala]

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

OK sir