JEE Main 2023 — Mathematics Calculus

Given,

$${\int }_{-0.15}^{0.15}|100{x}^{2}-1|dx=\frac{k}{3000}$$∫−0.150.15​∣100x2−1∣dx=3000k​

Now we know that, $\displaystyle {\int }_{-a}^{a}f(x)dx=2{\int }_{0}^{a}f(x)dx\text{if}f(x)\text{is even}$∫−aa​f(x)dx=2∫0a​f(x)dxiff(x)is even

So, $\displaystyle {\int }_{-0.15}^{0.15}|100{x}^{2}-1|dx=2{\int }_{0}^{0.15}|100{x}^{2}-1|dx$∫−0.150.15​∣100x2−1∣dx=2∫00.15​∣100x2−1∣dx

Now solving, $\displaystyle 2{\int }_{0}^{0.15}|100{x}^{2}-1|dx=2({\int }_{0}^{0.1}-(100{x}^{2}-1)dx+{\int }_{0.1}^{0.15}100{x}^{2}-1dx)$2∫00.15​∣100x2−1∣dx=2(∫00.1​−(100x2−1)dx+∫0.10.15​100x2−1dx)

$$\Rightarrow 2{\int }_{0}^{0.15}|100{x}^{2}-1|dx=2({[x-\frac{100{x}^{3}}{3}]}_{0}^{0.1}+{[\frac{100{x}^{3}}{3}-x]}_{0.1}^{0.15})$$⇒2∫00.15​∣100x2−1∣dx=2([x−3100x3​]00.1​+[3100x3​−x]0.10.15​)

$$\Rightarrow 2{\int }_{0}^{0.15}|100{x}^{2}-1|dx=2((\frac{1}{10}-\frac{1}{30})+(\frac{100\times {15}^{3}}{3\times {100}^{3}}-\frac{15}{100})-(\frac{1}{30}-\frac{1}{10}))$$⇒2∫00.15​∣100x2−1∣dx=2((101​−301​)+(3×1003100×153​−10015​)−(301​−101​))

$$\Rightarrow 2{\int }_{0}^{0.15}|100{x}^{2}-1|dx=2(\frac{2}{30}+(\frac{225}{2000}-\frac{15}{100})+\frac{2}{30})$$⇒2∫00.15​∣100x2−1∣dx=2(302​+(2000225​−10015​)+302​)

$$\Rightarrow 2{\int }_{0}^{0.15}|100{x}^{2}-1|dx=2(\frac{4}{30}+(\frac{-3}{80}))=(\frac{8}{30}-\frac{3}{40})=\frac{32-9}{120}=\frac{23}{120}=\frac{575}{3000}$$⇒2∫00.15​∣100x2−1∣dx=2(304​+(80−3​))=(308​−403​)=12032−9​=12023​=3000575​

$$\Rightarrow 2{\int }_{0}^{0.15}|100{x}^{2}-1|dx=\frac{575}{3000}$$⇒2∫00.15​∣100x2−1∣dx=3000575​

Now comparing with $\displaystyle {\int }_{-0.15}^{0.15}|100{x}^{2}-1|dx=\frac{k}{3000}$∫−0.150.15​∣100x2−1∣dx=3000k​, we get $k=575$k=575