Let Pk be a point on the curve y-sin2 x- whose x coordinate is k-n-1 k-1-2-3-n- If A is -1-0- then limn -1-n-k-1nAPk2-A-Bsin 2- then the value of 3A-B is

Lim_n →∞1n∑_k=1^n(( kn)^2+sin^2 ( kn 1)) =∫_0^1(x^2+sin^2 (x 1))dx =x^33+∫_0^1sin^2 (1 x 1)dx =x^23+ x2 12sin 2x2 =13+12 12sin22 =56 14sin 2 ⇒ A=56 and B=14 AB= ...

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Byju's Answer

Standard XII

Mathematics

Definite Integral as Limit of Sum

Let Pk be a p...

Question

Let $P_{k}$ be a point on the curve $y = {sin}^{2} x$ , whose $x$ coordinate is $\frac{k}{n} - 1 (k = 1, 2, 3, . . ., n)$ . If $A$ is $(- 1, 0)$ , then $lim n \to \infty \frac{1}{n} n \sum k = 1 (A P_{k})^{2} = A - B sin 2$ , then the value of $\frac{3 A}{B}$ is

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Solution

$lim n \to \infty \frac{1}{n} n \sum k = 1 ({(\frac{k}{n})}^{2} + {sin}^{2} (\frac{k}{n} - 1))$
$= \int_{0}^{1} (x^{2} + {sin}^{2} (x - 1)) d x$
$= \frac{x^{3}}{3} + \int_{0}^{1} {sin}^{2} (1 - x - 1) d x$
$= \frac{x^{2}}{3} + \frac{x}{2} - \frac{1}{2} \frac{sin 2 x}{2}$
$= \frac{1}{3} + \frac{1}{2} - \frac{1}{2} \frac{sin 2}{2}$
$= \frac{5}{6} - \frac{1}{4} sin 2$
$\Rightarrow A = \frac{5}{6}$ and $B = \frac{1}{4}$
$\frac{A}{B} = \frac{10}{3}$

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Let Pk be a point on the curve y=sin2x, whose x coordinate is kn−1 (k=1,2,3,...,n). If A is (−1,0), then limn→∞1nn∑k=1(APk)2=A−Bsin2, then the value of 3AB is

limn→∞1nn∑k=1((kn)2+sin2(kn−1)) =∫10(x2+sin2(x−1))dx =x33+∫10sin2(1−x−1)dx =x23+x2−12sin2x2 =13+12−12sin22 =56−14sin2 ⇒A=56 and B=14 AB=103

Let $P_{k}$ be a point on the curve $y = {sin}^{2} x$ , whose $x$ coordinate is $\frac{k}{n} - 1 (k = 1, 2, 3, . . ., n)$ . If $A$ is $(- 1, 0)$ , then $lim n \to \infty \frac{1}{n} n \sum k = 1 (A P_{k})^{2} = A - B sin 2$ , then the value of $\frac{3 A}{B}$ is