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MathHardAdv 2021 • Paper 2

Evaluation of Scaled Definite Integral for Absolute Value Weight Function

Comprehension Passage

Let gi:[π8,3π8]→Rg_i: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \to \mathbb{R}, i=1,2i = 1, 2, and f:[π8,3π8]→Rf: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \to \mathbb{R} be functions defined such that g1(x)=1,g2(x)=∣4x−π∣andf(x)=sin⁡2x,for all x∈[π8,3π8]g_1(x) = 1, \quad g_2(x) = |4x - \pi| \quad \text{and} \quad f(x) = \sin^2 x, \quad \text{for all } x \in \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] Define integrals Si=∫π83π8f(x)⋅gi(x)dx,i=1,2S_i = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) dx, \quad i = 1, 2

Find the value of 48S2π2\frac{48S_2}{\pi^2}.

Official Numerical Answer1.5

Topics & Concepts

IntegralsDefinite IntegrationProperties of Definite Integrals

Step-by-Step Solution

To find the value of 48S2π2\frac{48S_2}{\pi^2}, we first evaluate the integral S2S_2 given by: S2=∫π83π8f(x)⋅g2(x) dx=∫π83π8∣4x−π∣sin⁡2x dxS_2 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_2(x) \, dx = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} |4x - \pi| \sin^2 x \, dx

We can use the substitution method by performing a change of variable centered at the midpoint of the integration interval [π8,3π8]\left[\frac{\pi}{8}, \frac{3\pi}{8}\right], which is x=π4x = \frac{\pi}{4}.

Let x=π4+t  ⟹  dx=dtx = \frac{\pi}{4} + t \implies dx = dt.

The limits of integration transform as follows:

  • When x=π8x = \frac{\pi}{8}, t=π8−π4=−π8t = \frac{\pi}{8} - \frac{\pi}{4} = -\frac{\pi}{8}
  • When x=3π8x = \frac{3\pi}{8}, t=3π8−π4=π8t = \frac{3\pi}{8} - \frac{\pi}{4} = \frac{\pi}{8}

Now, substitute x=π4+tx = \frac{\pi}{4} + t into the terms inside the integral:

  1. ∣4x−π∣=∣4(π4+t)−π∣=∣4t∣=4∣t∣|4x - \pi| = \left|4\left(\frac{\pi}{4} + t\right) - \pi\right| = |4t| = 4|t|
  2. sin⁡2x=sin⁡2(π4+t)=1−cos⁡(2(π4+t))2=1−cos⁡(π2+2t)2=1+sin⁡(2t)2\sin^2 x = \sin^2\left(\frac{\pi}{4} + t\right) = \frac{1 - \cos\left(2\left(\frac{\pi}{4} + t\right)\right)}{2} = \frac{1 - \cos\left(\frac{\pi}{2} + 2t\right)}{2} = \frac{1 + \sin(2t)}{2}

Substituting these back into the integral for S2S_2: S2=∫−π8π84∣t∣⋅(1+sin⁡(2t)2)dtS_2 = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4|t| \cdot \left(\frac{1 + \sin(2t)}{2}\right) dt S2=2∫−π8π8∣t∣(1+sin⁡(2t)) dtS_2 = 2 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| (1 + \sin(2t)) \, dt S2=2∫−π8π8∣t∣ dt+2∫−π8π8∣t∣sin⁡(2t) dtS_2 = 2 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \, dt + 2 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \sin(2t) \, dt

Observe the properties of the integrands:

  • The function h1(t)=∣t∣sin⁡(2t)h_1(t) = |t| \sin(2t) is an odd function since h1(−t)=∣−t∣sin⁡(−2t)=−∣t∣sin⁡(2t)=−h1(t)h_1(-t) = |-t|\sin(-2t) = -|t|\sin(2t) = -h_1(t). Since the limits are symmetric about 00, we have: ∫−π8π8∣t∣sin⁡(2t) dt=0\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \sin(2t) \, dt = 0

  • The function h2(t)=∣t∣h_2(t) = |t| is an even function, so: ∫−π8π8∣t∣ dt=2∫0π8t dt\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \, dt = 2 \int_{0}^{\frac{\pi}{8}} t \, dt

Thus, the integral S2S_2 simplifies to: S2=2(2∫0π8t dt)+0S_2 = 2 \left( 2 \int_{0}^{\frac{\pi}{8}} t \, dt \right) + 0 S2=4[t22]0π8=2(π8)2=2×π264=π232S_2 = 4 \left[ \frac{t^2}{2} \right]_{0}^{\frac{\pi}{8}} = 2 \left( \frac{\pi}{8} \right)^2 = 2 \times \frac{\pi^2}{64} = \frac{\pi^2}{32}

Now, we calculate the required expression 48S2π2\frac{48S_2}{\pi^2}: 48S2π2=48π2×π232=4832=1.5\frac{48S_2}{\pi^2} = \frac{48}{\pi^2} \times \frac{\pi^2}{32} = \frac{48}{32} = 1.5

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