let u = a^2-x^2 and you have
du = -2x dx
So the integral becomes
∫[a,0] -1/2 √u du
= -1/3 u^(3/2) [a^2,0]
= 1/3 a^3
Evaluate the following integral:
(0)S(a)((x)((a^2 - x^2)^(1/2)))dx
with a being a constant and the (0) being at the bottom of the integral notation and (a) at the top. S is the integral notation.
I firstly checked whether the function was even/odd. (x) is odd, and ((a^2 - x^2)^(1/2)) is even, therefore the question answers 0. I am hesitant as to whether this is right because the professor did not do this and went straight to solving the question using substitution (and the outcome was not 0.)
I appreciate any help.
du = -2x dx
So the integral becomes
∫[a,0] -1/2 √u du
= -1/3 u^(3/2) [a^2,0]
= 1/3 a^3
That is not the case here.
The integral you have is:
∫[from 0 to a] x(a² - x²)^(1/2) dx.
To approach this integral, you have correctly identified that the function x is odd, and the function (a² - x²)^(1/2) is even. This suggests that the product of these two functions could result in an odd function.
Let's see how the substitution method can help give us a definitive answer:
1. Use the substitution: x = a sinθ.
- When x = 0, θ = 0.
- When x = a, θ = π/2.
2. Calculate dx in terms of dθ using the derivative of the substitution:
dx = a cosθ dθ.
3. Substitute x and dx in terms of θ into the integral:
∫[from 0 to π/2] a sinθ(a² - (a sinθ)²)^(1/2) (a cosθ) dθ.
4. Simplify the expression under the square root:
(a² - a² sin²θ)^(1/2) = a(1 - sin²θ)^(1/2) = a cosθ.
5. Rewrite the integral using the simplified expression:
∫[from 0 to π/2] a² sinθ cos²θ dθ.
Now, we can simplify further using a trigonometric identity. Recall that cos²θ = (1 + cos2θ)/2.
6. Rewrite the integral using the trigonometric identity:
∫[from 0 to π/2] a² sinθ (1 + cos2θ)/2 dθ.
7. Distribute the a²/2 term into the integral:
(a²/2) ∫[from 0 to π/2] sinθ + sinθ cos2θ dθ.
8. Integrate each term separately:
(a²/2) [-cosθ - (1/3)cos³θ] evaluated from 0 to π/2.
Now, let's evaluate the integral at the upper and lower limits:
- At θ = π/2:
-cos(π/2) - (1/3)cos³(π/2) = 0 + 0 = 0.
- At θ = 0:
-cos(0) - (1/3)cos³(0) = -1 - (1/3)(-1) = -1 + 1/3 = -2/3.
9. Calculate the difference between the upper and lower limits:
0 - (-2/3) = 2/3.
Therefore, the value of the integral is 2/3.
In conclusion, the integral ∫[from 0 to a] x(a² - x²)^(1/2) dx evaluates to 2/3, which means the answer is not zero. It appears that your initial assumption about the odd and even nature of the functions did not lead to the correct result.