Thread: 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 3π/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?

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

1) Yes, but I would write the result as instead.

2.) I would use a substitution to make the computation simpler: and write:

Originally Posted by arsenal12345

(1)


[/B]3π/4
∫ sins x dx
π/2

3π/4
∫ sins x dx = -cos x
π/2

(-cos 3π/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 should be written as , which is read as ' 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 using binomial expansion and evaluated the integral as in (2)

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 instead.

2.) I would use a substitution to make the computation simpler: and write:

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?

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.