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:
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:
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.
You can expand using binomial expansion and evaluated the integral as in (2)
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.