# Math Help - intigration! some1 please cheack my work and help me with a problem!

1. ## intigration! some1 please cheack my work and help me with a problem!

how do i do this?

∫ (x-1)^2 x dx

are these right?
(1)

3π/4
∫ sins x dx
π/2

3π/4
∫ sins x dx = -cos x
π/2

(-cos 3
π/4) - (-cos π/2 ) = 0.71

(2)

2
∫ (2+x) dx
0

2
∫ (2+x) dx =
2x+ (1/2)x^2
0

(2(2)+(1/2)(2)^2) - ( 2(0) + (1/2)(0)^2)

=6
are these right?

2. ## Re: intigration! some1 please cheack my work and help me with a problem!

For problem 0) expand the binomial, then distribute the x to get a polynomial integrand.

1) Yes, but I would write the result as $\frac{1}{\sqrt{2}}$ instead.

2.) I would use a substitution to make the computation simpler: $u=x+2,\:du=dx$ and write:

$\int_2^4 u\,du=\frac{1}{2}\left(4^2-2^2 \right)=6$

3. ## Re: intigration! some1 please cheack my work and help me with a problem!

Originally Posted by arsenal12345
(1)

[/B]3π/4
∫ sins x dx
π/2

3π/4
∫ sins x dx = -cos x
π/2

(-cos 3
π/4) - (-cos π/2 ) = 0.71

(2)

2
∫ (2+x) dx
0

2
∫ (2+x) dx =
2x+ (1/2)x^2
0

(2(2)+(1/2)(2)^2) - ( 2(0) + (1/2)(0)^2)

=6
are these right?
Your solution is correct. However your notation of writing $\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin x dx = -\cos x$ should be written as $\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin x dx = -\cos x \vert_{\frac{\pi}{2}} ^{\frac{3\pi}{4}}$, which is read as ' $-\cos x$ evaluated at the lower limit and higher limit of integration'. Use the same notation in the second integral as well.

Originally Posted by arsenal12345
how do i do this?
∫ (x-1)^2 x dx
You can expand $(x-1)^2$ using binomial expansion and evaluated the integral as in (2)

4. ## Re: intigration! some1 please cheack my work and help me with a problem!

Originally Posted by MarkFL2
For problem 0) expand the binomial, then distribute the x to get a polynomial integrand.

1) Yes, but I would write the result as $\frac{1}{\sqrt{2}}$ instead.

2.) I would use a substitution to make the computation simpler: $u=x+2\:du=dx$ and write:

$\int_2^4 u\,du=\frac{1}{2}\left(4^2-2^2 \right)=6$
is my working right though? would get full marks for what i did without making the changes mentioned?

also regarding the first one is it

(x^2-2x+1)x = x^3-2x^2+3

∫x^3-2x^2+3

(1/4)x^4 -(2/3)x^3+(1/2)(x^2)+c

is this right? with proper working?

5. ## Re: intigration! some1 please cheack my work and help me with a problem!

You expanded the binomial correctly, but did not distribute the x correctly, although your final answer is correct...

I am not a teacher, but I would not give full credit for 0.71 when it may easily be given exactly.