JEE Challenger
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MathMedAdv 2020 • Paper 1

Estimating Definite Integrals Involving Trigonometric Functions

Which of the following integral inequalities is/are TRUE?

Options

A

∫01xcos⁡x dx≥38\int_0^1 x \cos x \, dx \geq \frac{3}{8}

Correct
B

∫01xsin⁡x dx≥310\int_0^1 x \sin x \, dx \geq \frac{3}{10}

Correct
C

∫01x2cos⁡x dx≥12\int_0^1 x^2 \cos x \, dx \geq \frac{1}{2}

D

∫01x2sin⁡x dx≥29\int_0^1 x^2 \sin x \, dx \geq \frac{2}{9}

Correct

Topics & Concepts

IntegralsDefinite IntegrationProperties of Definite Integrals

Step-by-Step Solution

To determine the correct integral inequalities, we can use standard Taylor series expansion inequalities for sin⁡x\sin x and cos⁡x\cos x on the interval x∈[0,1]x \in [0, 1].

Useful Inequalities for x∈[0,1]x \in [0, 1]:

  1. Proof for cos⁡x≥1−x22\cos x \geq 1 - \frac{x^2}{2}: Let f(x)=cos⁡x−1+x22f(x) = \cos x - 1 + \frac{x^2}{2}. The derivative is f′(x)=−sin⁡x+x≥0f'(x) = -\sin x + x \geq 0 for all x≥0x \geq 0 since sin⁡x≤x\sin x \leq x. Since f(0)=0f(0) = 0 and f′(x)≥0f'(x) \geq 0 for x∈[0,1]x \in [0, 1], f(x)≥0f(x) \geq 0, which implies: cos⁡x≥1−x22\cos x \geq 1 - \frac{x^2}{2}

  2. Proof for sin⁡x≥x−x36\sin x \geq x - \frac{x^3}{6}: Let g(x)=sin⁡x−x+x36g(x) = \sin x - x + \frac{x^3}{6}. The derivative is g′(x)=cos⁡x−1+x22=f(x)≥0g'(x) = \cos x - 1 + \frac{x^2}{2} = f(x) \geq 0. Since g(0)=0g(0) = 0 and g′(x)≥0g'(x) \geq 0 for x∈[0,1]x \in [0, 1], g(x)≥0g(x) \geq 0, which implies: sin⁡x≥x−x36\sin x \geq x - \frac{x^3}{6}

Evaluating Each Option:

Option (A):

Using cos⁡x≥1−x22\cos x \geq 1 - \frac{x^2}{2} for x∈[0,1]x \in [0, 1]: xcos⁡x≥x−x32x \cos x \geq x - \frac{x^3}{2}

Integrating both sides from 00 to 11: ∫01xcos⁡x dx≥∫01(x−x32)dx=[x22−x48]01=12−18=38\int_0^1 x \cos x \, dx \geq \int_0^1 \left(x - \frac{x^3}{2}\right) dx = \left[ \frac{x^2}{2} - \frac{x^4}{8} \right]_0^1 = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}

Thus, Option (A) is TRUE.

Option (B):

Using sin⁡x≥x−x36\sin x \geq x - \frac{x^3}{6} for x∈[0,1]x \in [0, 1]: xsin⁡x≥x2−x46x \sin x \geq x^2 - \frac{x^4}{6}

Integrating both sides from 00 to 11: ∫01xsin⁡x dx≥∫01(x2−x46)dx=[x33−x530]01=13−130=930=310\int_0^1 x \sin x \, dx \geq \int_0^1 \left(x^2 - \frac{x^4}{6}\right) dx = \left[ \frac{x^3}{3} - \frac{x^5}{30} \right]_0^1 = \frac{1}{3} - \frac{1}{30} = \frac{9}{30} = \frac{3}{10}

Thus, Option (B) is TRUE.

Option (C):

Since cos⁡x≤1\cos x \leq 1 for all x∈[0,1]x \in [0, 1]: x2cos⁡x≤x2x^2 \cos x \leq x^2

Integrating both sides from 00 to 11: ∫01x2cos⁡x dx≤∫01x2 dx=[x33]01=13\int_0^1 x^2 \cos x \, dx \leq \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}

Since 13<12\frac{1}{3} < \frac{1}{2}, it follows that: ∫01x2cos⁡x dx<12\int_0^1 x^2 \cos x \, dx < \frac{1}{2}

Thus, Option (C) is FALSE.

Option (D):

Using sin⁡x≥x−x36\sin x \geq x - \frac{x^3}{6} for x∈[0,1]x \in [0, 1]: x2sin⁡x≥x3−x56x^2 \sin x \geq x^3 - \frac{x^5}{6}

Integrating both sides from 00 to 11: ∫01x2sin⁡x dx≥∫01(x3−x56)dx=[x44−x636]01=14−136=836=29\int_0^1 x^2 \sin x \, dx \geq \int_0^1 \left(x^3 - \frac{x^5}{6}\right) dx = \left[ \frac{x^4}{4} - \frac{x^6}{36} \right]_0^1 = \frac{1}{4} - \frac{1}{36} = \frac{8}{36} = \frac{2}{9}

Thus, Option (D) is TRUE.

Conclusion:

The correct statements are A, B, and D.

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