# Integrals help

1. Apr 5, 2014

### Mathsishard123

I have these integrals to find:

∫ (5x^2 + sqrt(x) - 4/x^2) dx

∫ [cos(x/2) - sin(3x/2)] dx

∫ s/sqrt(s^2 + 4) ds (upper coordinate is 5 lower coordinate is 1)

I have worked it out as:

∫〖(5x^2+√x〗-4/x^2) dx=5x^3/(2+1)+x^(1/2+1)/(1+1/2)-4x^(-2+1)/(-2+1)+C=5/3 x^3+2/3x^(3/2)+4/x+C

∫〖(cos⁡(x/2)-sin⁡(3x/2) )dx=2 sin⁡(x/2)+2/3 cos⁡(3x/2)+C〗

∫_1^5〖s/√(s^2+4)〗ds=1/2 √(s^2+4) (1≤s≤5)=1/2 (5^2+1)^(1/2)-1/2(1^2+1)^(1/2)=√26/2-√2/2

Do these look right?

2. Apr 5, 2014

### Panphobia

The first one is kind of messy but you know you can split integrals like that up?
∫ (5x^2 + sqrt(x) - 4/x^2) dx = ∫5x^2 dx + ∫sqrt(x) dx - ∫4/x^2 dx
That should make it easier. The second integral is right.
By coordinate, do you mean bounds in a definite integral? so ∫ s/sqrt(s^2 + 4) ds = sqrt(s^2 + 4) + C
sqrt(5^2+4)-sqrt(1^2+4), so sqrt(29) - sqrt(5) should be the last one.

Last edited: Apr 6, 2014
3. Apr 6, 2014

### Staff: Mentor

No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.

4. Apr 6, 2014

### Panphobia

Look
u = s^2 + 4
du = 2s ds

(1/2)∫du/sqrt(u)

(2/2) sqrt(u)

sqrt(s^2 + 4) + C
Where am I missing the factor of 1/2?

5. Apr 6, 2014

### ehild

You missed a factor of 2.

ehild

6. Apr 6, 2014

### Curious3141

Panphobia is correct.

7. Apr 6, 2014

### Staff: Mentor

I shouldn't do integration in my head.

8. Apr 6, 2014

### Curious3141

Why not? I did, and it was right. :tongue:

Just messin' with you.