If $\int_{-0.15}^{0.15}\left|100 \mathrm{x}^{2}-1\right| \mathrm{dx}=\frac{\mathrm{k}}{3000}$, then k is equal to ______.

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Definite Integrals

\int_{- 0.15}^{0.15} 100 x^{2} - 1 dx = \frac{k}{3000}

, then k is equal to ______.

Answer: 575

The integrand

∣100 x^{2} - 1∣

is even, so

\int_{- 0.15}^{0.15} ∣100 x^{2} - 1∣ d x = 2 \int_{0}^{0.15} ∣100 x^{2} - 1∣ d x

100 x^{2} - 1 = 0

x = \frac{1}{10}

. For

0 \leq x < \frac{1}{10}

100 x^{2} - 1 < 0

; for

\frac{1}{10} < x \leq \frac{3}{20}

100 x^{2} - 1 > 0

I = 2 [\int_{0}^{1/10} (1 - 100 x^{2}) d x + \int_{1/10}^{3/20} (100 x^{2} - 1) d x]

First part:

[x - \frac{100 x ^{3}}{3}]_{0}^{1/10} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}

Second part:

[\frac{100 x ^{3}}{3} - x]_{1/10}^{3/20} = (\frac{9}{80} - \frac{3}{20}) - (\frac{1}{30} - \frac{1}{10}) = - \frac{3}{80} + \frac{1}{15} = \frac{- 9 + 16}{240} = \frac{7}{240}

∴ I = 2 (\frac{1}{15} + \frac{7}{240}) = 2 (\frac{16 + 7}{240}) = 2 \cdot \frac{23}{240} = \frac{23}{120}

Given

I = \frac{k}{3000}

\frac{23}{120} = \frac{k}{3000} \Rightarrow k = \frac{3000 \times 23}{120} = 25 \times 23 = 575

Hence

k = 575

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer