Finding Integrals: ∫ (5x^2 + sqrt(x) - 4/x^2) dx

In summary: No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.
  • #1
Mathsishard123
2
0
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?
 
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  • #2
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:
  • #3
Panphobia said:
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
No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.
Panphobia said:
sqrt(5^2+4)-sqrt(1^2+4), so sqrt(29) - sqrt(5) should be the last one.
 
  • #4
Mark44 said:
No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.
Are you talking about this? ∫ s/sqrt(s^2 + 4) ds = sqrt(s^2 + 4) + C
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
Mathsishard123 said:
∫_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?

You missed a factor of 2.

ehild
 
  • #6
Mark44 said:
No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.

Panphobia is correct.
 
  • #8
Mark44 said:
I shouldn't do integration in my head.:cry:

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

Just messin' with you. :biggrin:
 

Related to Finding Integrals: ∫ (5x^2 + sqrt(x) - 4/x^2) dx

1. What is the process for finding the integral of a polynomial expression?

The process for finding the integral of a polynomial expression is to use the power rule, which states that the integral of x^n is (x^(n+1))/(n+1), where n is the power of x. In this case, the integral of 5x^2 would be (5x^3)/3.

2. How do you handle square root expressions when finding an integral?

To handle square root expressions, you can use the substitution method. Let u be equal to the expression inside the square root, then replace the square root with u and solve for u. In this case, u would be equal to x, and the integral of sqrt(x) would be (2/3)x^(3/2).

3. Is there a specific technique for finding the integral of a fraction?

Yes, you can use the inverse power rule, which states that the integral of x^n is (x^(n+1))/(n+1). In this case, the integral of 4/x^2 would be -4/x.

4. Can the integral of a function be negative?

Yes, the integral of a function can be negative. The integral represents the area under the curve, so if the curve is below the x-axis, the area will be negative.

5. Are there any other methods for finding integrals besides the ones mentioned?

Yes, there are other methods such as integration by parts, trigonometric substitution, and partial fractions. These methods are used for more complex functions that cannot be solved using the power rule or substitution method.

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