Integral Calculus – Definite Integration – JEE Main 28 Jan 2026 Shift 1

Solution:

Let $I_r = \int_{0}^{r}x|\sin \pi x|dx$.
We can use the property $\int_0^r f(x) dx = \int_0^r (r-x) f(r-x) dx$. However, $|\sin \pi(r-x)| = |\sin(\pi r – \pi x)| = |\sin \pi x|$ for integer $r$.
$$I_r = \int_0^r (r-x) |\sin \pi x| dx = r\int_0^r |\sin \pi x| dx – I_r$$
$$2I_r = r \int_0^r |\sin \pi x| dx$$

Since $|\sin \pi x|$ is periodic with period 1:
$$\int_0^r |\sin \pi x| dx = r \int_0^1 |\sin \pi x| dx$$
Evaluate $\int_0^1 \sin \pi x dx$ (positive in first quadrant):
$$= \left[ -\frac{\cos \pi x}{\pi} \right]_0^1 = -\frac{1}{\pi}(-1 – 1) = \frac{2}{\pi}$$

Substitute back:
$$2I_r = r \cdot r \left(\frac{2}{\pi}\right) \Rightarrow I_r = \frac{r^2}{\pi}$$

Now evaluate the term in the summation:
$$\sqrt{\pi I_r} = \sqrt{\pi \cdot \frac{r^2}{\pi}} = \sqrt{r^2} = r$$

The summation becomes:
$$\sum_{r=1}^{20} r = \frac{20(21)}{2} = 10(21) = 210$$

Ans. 210