How do you find the definite integral for: #(9-x^2)^(1/2)# for the intervals #[-3, 3]#?

Question

Ezra Adams · Accepted Answer

The answer is #=(9pi)/2# First, we calculate #intsqrt(9-x^2)dx# let #x=3sinu#, then #dx=3cosudu# #sqrt(9-9sin^2u)=3cosu# #:.intsqrt(9-x^2)dx=int3cosu*3cosudu# #=9intcos^2udu# #cos2u=2cos^2u-1##=>##cos^2u=(cos2u+1)/2# #9intcos^2udu=9/2int(cos2u+1)du# #=9/2((sin2u)/2+u)= 9/4((sin2u)+2u)# #sin2u=2sinucosu=2*x/3*sqrt(9-x^2)/3=2/9x(sqrt(9-x^2))# #9/4((sin2u)+2u)=9/4*2/9*(xsqrt(9-x^2))+9/2arcsin(x/3)# #int_-3^3sqrt(9-x^2)dx=(1/2(xsqrt(9-x^2))+9/2arcsin(x/3))_-3^3# #0+9/2*arcsin1-0-9/2arcsin(-1)# #0+9pi/4--9pi/4=9pi/2#

Elijah Abram · Answer

#(9pi)/2# We can also think about this geometrically. #y=sqrt(9-x^2)# is the upper half of the circle #x^2+y^2=9#, a circle with radius #3# and center at #(0,0)#. Since the bounds span the length of the semicircle, the integral will be equal to #1//2# the circle's area, or #(9pi)/2#.

Christopher Agresta · Answer

undefined To find the definite integral of $(9-x^2)^{\frac{1}{2}}$ over the interval $[-3, 3]$, we can use the formula for the definite integral of a function. The definite integral of $f(x)$ from $a$ to $b$ is given by:

$$\int_{a}^{b} f(x) \, dx$$

Substituting $f(x) = (9-x^2)^{\frac{1}{2}}$ and the interval $[-3, 3]$, we have:

$$\int_{-3}^{3} (9-x^2)^{\frac{1}{2}} \, dx$$

To evaluate this integral, we can use trigonometric substitution. Let $x = 3\sin(\theta)$, then $dx = 3\cos(\theta) \, d\theta$. When $x = -3$, $\theta = -\frac{\pi}{2}$, and when $x = 3$, $\theta = \frac{\pi}{2}$. So, the integral becomes:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (9-(3\sin(\theta))^2)^{\frac{1}{2}} \cdot 3\cos(\theta) \, d\theta$$

Simplify the integrand:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (9-9\sin^2(\theta))^{\frac{1}{2}} \cdot 3\cos(\theta) \, d\theta$$
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (9\cos^2(\theta))^{\frac{1}{2}} \cdot 3\cos(\theta) \, d\theta$$
$$3 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3\cos^2(\theta) \, d\theta$$

Now, use the double angle identity for cosine, $\cos^2(\theta) = \frac{1}{2}(1+\cos(2\theta))$:

$$3 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1}{2}(1+\cos(2\theta)) \right) \, d\theta$$
$$3 \left[ \frac{1}{2}\left(\theta + \frac{1}{2}\sin(2\theta)\right) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$$

Now, evaluate the definite integral:

$$3 \left[ \frac{1}{2}\left(\frac{\pi}{2} + \frac{1}{2}\sin(\pi)\right) - \frac{1}{2}\left(-\frac{\pi}{2} + \frac{1}{2}\sin(-\pi)\right) \right]$$
$$3 \left[ \frac{1}{2}\left(\frac{\pi}{2}\right) - \frac{1}{2}\left(-\frac{\pi}{2}\right) \right]$$
$$3 \left[ \frac{\pi}{4} + \frac{\pi}{4} \right]$$
$$3 \cdot \frac{\pi}{2}$$
$$\frac{3\pi}{2}$$

Therefore, the definite integral of $(9-x^2)^{\frac{1}{2}}$ over the interval $[-3, 3]$ is $\frac{3\pi}{2}$.

Paisley Johnson · Answer

undefined To find the definite integral of (9-x^2)^(1/2) for the interval [-3, 3], you can use the definite integral formula for a continuous function over a closed interval. In this case, you'll use the substitution method.

1. Make the substitution: Let $ x = 3\sin(\theta) $, then $ dx = 3\cos(\theta) d\theta $.
2. Adjust the limits of integration: When $ x = -3 $, $ \theta = -\frac{\pi}{2} $, and when $ x = 3 $, $ \theta = \frac{\pi}{2} $.
3. Rewrite the integral in terms of $ \theta $: $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (9 - (3\sin(\theta))^2)^{\frac{1}{2}} \cdot 3\cos(\theta) d\theta $.
4. Simplify the integrand: $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (9 - 9\sin^2(\theta))^{\frac{1}{2}} \cdot 3\cos(\theta) d\theta $.
5. Use the trigonometric identity: $ \sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta)) $ to simplify the integrand.
6. Substitute the identity into the integral and simplify further.
7. Evaluate the integral from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $.

The resulting value will be the definite integral of $ (9-x^2)^\frac{1}{2} $ over the interval [-3, 3].