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MathMedAdv 2024 • Paper 2

Evaluate Scaled Definite Integral of Product of Functions

Comprehension Passage

Let f:[0,π2]→[0,1]f : \left[0, \frac{\pi}{2}\right] \rightarrow [0, 1] be the function defined by f(x)=sin⁡2xf(x) = \sin^2 x and let g:[0,π2]→[0,∞)g : \left[0, \frac{\pi}{2}\right] \rightarrow [0, \infty) be the function defined by g(x)=πx2−x2g(x) = \sqrt{\frac{\pi x}{2} - x^2}.

The value of 16π3∫0π2f(x)g(x)dx\frac{16}{\pi^3} \int_{0}^{\frac{\pi}{2}} f(x) g(x) dx is ________.

Official Numerical Answer0.25

Topics & Concepts

IntegralsDefinite Integration

Step-by-Step Solution

To evaluate the given definite integral, let the integral be denoted as:
I=∫0π2f(x)g(x) dx=∫0π2sin⁡2xπx2−x2 dxI = \int_{0}^{\frac{\pi}{2}} f(x) g(x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin^2 x \sqrt{\frac{\pi x}{2} - x^2} \, dx

Step 1: Analyze the symmetry of g(x)g(x)

Using the King's Property of definite integrals, ∫abh(x) dx=∫abh(a+b−x) dx\int_{a}^{b} h(x) \, dx = \int_{a}^{b} h(a+b-x) \, dx, we replace xx with (π2−x)\left(\frac{\pi}{2} - x\right) in g(x)g(x):

g(π2−x)=π2(π2−x)−(π2−x)2g\left(\frac{\pi}{2} - x\right) = \sqrt{\frac{\pi}{2}\left(\frac{\pi}{2} - x\right) - \left(\frac{\pi}{2} - x\right)^2} g(π2−x)=π24−πx2−(π24−πx+x2)g\left(\frac{\pi}{2} - x\right) = \sqrt{\frac{\pi^2}{4} - \frac{\pi x}{2} - \left(\frac{\pi^2}{4} - \pi x + x^2\right)} g(π2−x)=πx2−x2=g(x)g\left(\frac{\pi}{2} - x\right) = \sqrt{\frac{\pi x}{2} - x^2} = g(x)

Thus, g(x)g(x) is symmetric about x=π4x = \frac{\pi}{4}.

Step 2: Apply King's Property to II

Using x→π2−xx \to \frac{\pi}{2} - x:
I=∫0π2sin⁡2(π2−x)g(π2−x)dx=∫0π2cos⁡2x g(x) dxI = \int_{0}^{\frac{\pi}{2}} \sin^2\left(\frac{\pi}{2} - x\right) g\left(\frac{\pi}{2} - x\right) dx = \int_{0}^{\frac{\pi}{2}} \cos^2 x \, g(x) \, dx

Adding the two representations of II:
2I=∫0π2(sin⁡2x+cos⁡2x)g(x) dx=∫0π2g(x) dx2I = \int_{0}^{\frac{\pi}{2}} \left(\sin^2 x + \cos^2 x\right) g(x) \, dx = \int_{0}^{\frac{\pi}{2}} g(x) \, dx

Therefore, I=12∫0π2g(x) dxI = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} g(x) \, dx

Step 3: Compute the integral of g(x)g(x)

The integrand g(x)g(x) can be rewritten by completing the square:
g(x)=πx2−x2=(π4)2−(x−π4)2g(x) = \sqrt{\frac{\pi x}{2} - x^2} = \sqrt{\left(\frac{\pi}{4}\right)^2 - \left(x - \frac{\pi}{4}\right)^2}

Let J=∫0π2(π4)2−(x−π4)2dxJ = \int_{0}^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2 - \left(x - \frac{\pi}{4}\right)^2} dx.

Using the substitution u=x−π4u = x - \frac{\pi}{4}, du=dxdu = dx:

  • When x=0x = 0, u=−π4u = -\frac{\pi}{4}
  • When x=π2x = \frac{\pi}{2}, u=π4u = \frac{\pi}{4}

J=∫−π4π4(π4)2−u2 duJ = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{\left(\frac{\pi}{4}\right)^2 - u^2} \, du

This represents the area of a semicircle with radius R=π4R = \frac{\pi}{4}:
J=12πR2=12π(π4)2=π332J = \frac{1}{2} \pi R^2 = \frac{1}{2} \pi \left(\frac{\pi}{4}\right)^2 = \frac{\pi^3}{32}

Step 4: Calculate the final value

Now substituting JJ back into the relation for II:
I=12J=12×π332=π364I = \frac{1}{2} J = \frac{1}{2} \times \frac{\pi^3}{32} = \frac{\pi^3}{64}

We need to evaluate:
16π3∫0π2f(x)g(x) dx=16π3×I=16π3×π364=1664=14=0.25\frac{16}{\pi^3} \int_{0}^{\frac{\pi}{2}} f(x)g(x) \, dx = \frac{16}{\pi^3} \times I = \frac{16}{\pi^3} \times \frac{\pi^3}{64} = \frac{16}{64} = \frac{1}{4} = 0.25

The value of 16π3∫0π2f(x)g(x)dx\frac{16}{\pi^3} \int_{0}^{\frac{\pi}{2}} f(x) g(x) dx is 0.250.25 (or 14\frac{1}{4}).

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