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Q.Evaluate ∫₀² x² dx as the limit of a sum.

Kerala DhseKerala DHSE Plus Two Board 2022Subjective· 3mImportance★★★★★
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Partition [0,2] into n equal strips of width h = 2/n, form the Riemann sum, and take the limit as n → ∞.

By definition, ∫abf(x) dx=lim⁡n→∞h∑r=0n−1f(a+rh)\displaystyle\int_a^b f(x)\,dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1} f(a+rh), where h=b−anh=\dfrac{b-a}{n}.

Here a=0, b=2, f(x)=x2a=0,\ b=2,\ f(x)=x^2, so h=2nh=\dfrac{2}{n} and a+rh=rha+rh = rh.

∫02x2 dx=lim⁡n→∞h∑r=0n−1(rh)2=lim⁡n→∞h3∑r=0n−1r2\displaystyle\int_0^2 x^2\,dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1}(rh)^2 = \lim_{n\to\infty} h^3\sum_{r=0}^{n-1}r^2

Using ∑r=0n−1r2=(n−1)n(2n−1)6\displaystyle\sum_{r=0}^{n-1}r^2 = \dfrac{(n-1)n(2n-1)}{6} and h3=8n3h^3=\dfrac{8}{n^3}:

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