The value of $\sum_{\mathrm{r}=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{\mathrm{r}} \mathrm{x} | \sin \pi \mathrm{x} | \mathrm{dx}\right)}\right|\right)$ is ______.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Definite Integrals

The value of

\sum_{r = 1}^{20} (π (\int_{0}^{r} x ∣ sin π x ∣ dx))

is ______.

Answer: 210

Let

I_{r} = \int_{0}^{r} x ∣ sin π x ∣ d x

. Replacing

x

r - x

and using that

∣ sin π x ∣

has period

1

with

r

an integer gives

I_{r} = \int_{0}^{r} (r - x) ∣ sin π x ∣ d x

; adding the two forms,

2 I_{r} = r \int_{0}^{r} ∣ sin π x ∣ d x = r \cdot \frac{2 r}{π}

, so

I_{r} = \frac{r ^{2}}{π}

. Then

π I_{r} = r^{2} = r

, and the required sum is

\sum_{r = 1}^{20} r = \frac{20 \cdot 21}{2} = 210

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer

The value of $\int_{- π /6}^{π /6} (\frac{π + 4 x ^{11}}{1 - s i n ( ∣ x ∣ + π /6 )}) d x$ is equal to
JEE Main 2026View solution →
$6 \int_{0}^{π} ∣ (sin 3 x + sin 2 x + sin x) ∣ d x$ is equal to.......
JEE Main 2026View solution →
Let $[\cdot]$ denote the greatest integer function and $f (x) = lim_{n \to \infty} \frac{1}{n ^{3}} \sum_{k = 1}^{n} [\frac{k ^{2}}{3 ^{x}}]$ . Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal …
JEE Main 2026View solution →
If $\int_{0}^{1} 4 cot^{- 1} (1 - 2 x + 4 x^{2}) dx = a tan^{- 1} (2) - b lo g_{e} (5)$ , where $a, b \in N$ , then $(2 a + b)$ is equal to ____.
JEE Main 2026View solution →
Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is
JEE Main 2026View solution →
The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{[ x ] + 4}) dx$ , where [•] denotes the greatest integer function, is
JEE Main 2026View solution →