Math Problem Statement

816 ⁣x16xx2dx\int_8^{16}\!x\sqrt{16x-x^2}\,\mathrm{d}x

Solution

To solve the integral 816x16xx2dx\int_8^{16} x \sqrt{16x - x^2} \, dx, we'll proceed by simplifying the integrand using a trigonometric substitution. Here's the detailed process:

Step 1: Simplifying the Expression

First, recognize that the expression 16xx216x - x^2 can be rewritten as: 16xx2=(x216x)=(x8)2+64.16x - x^2 = - (x^2 - 16x) = -(x - 8)^2 + 64. Thus, we have: 16xx2=64(x8)2.\sqrt{16x - x^2} = \sqrt{64 - (x - 8)^2}. This is the equation of a semicircle with radius 8, centered at x=8x = 8. Therefore, trigonometric substitution is a useful approach.

Step 2: Substitution

Let’s make the substitution: x8=8sinθso thatdx=8cosθdθ.x - 8 = 8 \sin \theta \quad \text{so that} \quad dx = 8 \cos \theta \, d\theta. Thus, x=8(1+sinθ)x = 8(1 + \sin \theta), and the limits of integration need to be adjusted. When x=8x = 8, θ=0\theta = 0; and when x=16x = 16, θ=π2\theta = \frac{\pi}{2}.

Step 3: Substituting into the Integral

The square root term becomes: 64(x8)2=6464sin2θ=8cosθ.\sqrt{64 - (x - 8)^2} = \sqrt{64 - 64\sin^2\theta} = 8\cos\theta. Substitute into the original integral: 816x16xx2dx=0π28(1+sinθ)8cosθ8cosθdθ.\int_8^{16} x \sqrt{16x - x^2} \, dx = \int_0^{\frac{\pi}{2}} 8(1 + \sin\theta) \cdot 8 \cos\theta \cdot 8 \cos\theta \, d\theta. Simplifying this expression gives: 5120π2(1+sinθ)cos2θdθ.512 \int_0^{\frac{\pi}{2}} (1 + \sin\theta) \cos^2\theta \, d\theta.

Step 4: Solving the Integral

Expand the integrand: 5120π2(cos2θ+sinθcos2θ)dθ.512 \int_0^{\frac{\pi}{2}} (\cos^2\theta + \sin\theta \cos^2\theta) \, d\theta. Now solve these two integrals separately.

Integral 1: 0π2cos2θdθ\int_0^{\frac{\pi}{2}} \cos^2\theta \, d\theta

Use the identity cos2θ=1+cos2θ2\cos^2\theta = \frac{1 + \cos 2\theta}{2}: 0π2cos2θdθ=120π2(1+cos2θ)dθ=12[π2+0]=π4.\int_0^{\frac{\pi}{2}} \cos^2\theta \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 + \cos 2\theta) \, d\theta = \frac{1}{2} \left[ \frac{\pi}{2} + 0 \right] = \frac{\pi}{4}.

Integral 2: 0π2sinθcos2θdθ\int_0^{\frac{\pi}{2}} \sin\theta \cos^2\theta \, d\theta

Use the substitution u=cosθu = \cos\theta, du=sinθdθdu = -\sin\theta \, d\theta, so the integral becomes: 0π2sinθcos2θdθ=10u2du=01u2du=13.\int_0^{\frac{\pi}{2}} \sin\theta \cos^2\theta \, d\theta = \int_1^0 -u^2 \, du = \int_0^1 u^2 \, du = \frac{1}{3}.

Step 5: Final Answer

Now, combine the results: 512(π4+13)=512(3π+412)=512(3π+4)12=128(3π+4)3.512 \left( \frac{\pi}{4} + \frac{1}{3} \right) = 512 \left( \frac{3\pi + 4}{12} \right) = \frac{512(3\pi + 4)}{12} = \frac{128(3\pi + 4)}{3}.

Thus, the value of the integral is: 128(3π+4)3.\boxed{\frac{128(3\pi + 4)}{3}}.


Would you like further details on any of the steps, or have any questions?

Here are some related questions:

  1. How does trigonometric substitution help in evaluating integrals?
  2. What are the common forms of integrals involving square roots of quadratic expressions?
  3. What is the geometric interpretation of integrals involving circles or semicircles?
  4. How can the cosine double-angle identity simplify integrals?
  5. What other types of integrals can be solved using trigonometric substitution?

Tip: Trigonometric substitution is particularly useful when dealing with integrals involving a2x2\sqrt{a^2 - x^2}, a2+x2\sqrt{a^2 + x^2}, or x2a2\sqrt{x^2 - a^2}.

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Math Problem Analysis

Mathematical Concepts

Trigonometric Substitution
Integration
Geometry of Circles
Integral Calculus

Formulas

Trigonometric substitution: x - a = Rsin(θ)
Double-angle identity: cos^2(θ) = (1 + cos(2θ)) / 2

Theorems

Trigonometric substitution
Pythagorean identity: sin^2(θ) + cos^2(θ) = 1

Suitable Grade Level

College level (Calculus II or higher)