JEE Main 2023 (Online) 12th April Morning ShiftQuestion 1
If $\int_{-0.15}^{0.15}\left|100x^{2}-1\right|dx=\frac{k}{3000}$, then $k$ is equal to $\_\_\_\_\_\_$ .
Definite Integration
Answer$575$
Solution
Concept. $100x^2-1$ changes sign at $x=\pm\frac1{10}$. The integrand is even, so work on $[0,0.15]$ and double. Formulas used. (1) $|100x^2-1|=1-100x^2$ for $|x|\le\frac1{10}$, and $100x^2-1$ for $\frac1{10}\le|x|\le0.15$(2) $\int_{-a}^ag=2\int_0^ag$ for even $g$Step 1 - Split. By (1), (2):(Horizontal scale stretched.) $$I=2\Big[\int_0^{1/10}(1-100x^2)\,dx+\int_{1/10}^{3/20}(100x^2-1)\,dx\Big].$$ Step 2 - First part. $\frac1{10}-\frac{100}3\cdot\frac1{1000}=\frac1{10}-\frac1{30}=\frac1{15}$. Step 3 - Second part. $$\frac{100}3\Big(\frac{27}{8000}-\frac1{1000}\Big)-\Big(\frac3{20}-\frac1{10}\Big)=\frac{19}{240}-\frac1{20}=\frac7{240}.$$ Step 4 - Total. $I=2\Big(\frac{16}{240}+\frac7{240}\Big)=\frac{23}{120}=\frac{575}{3000}$. So $k=575$. Answer: 575