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Q.(a) What is the value of the definite integral of x(1 - x)^9 dx from 0 to 1 ?

(i) 1/10
(ii) 1/11
(iii) 1/90
(iv) 1/110 (Score : 1)
(b) Find the definite integral of (2x + 3) dx from 0 to 1 as the limit of a sum. (Scores : 3)
Kerala DhseKerala DHSE Plus Two Board 2015Subjective· 4mImportance★★★★★
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(a) substitute u=1−xu=1-x to turn it into a standard Beta-type integral. (b) build the Riemann sum with nn equal strips and take the limit.

(a) ∫01x(1−x)9dx\displaystyle\int_0^1 x(1-x)^9dx. Let u=1−xu=1-x (x=1−ux=1-u, dx=−dudx=-du; limits flip):

=∫01(1−u)u9 du=∫01(u9−u10)du=[u1010−u1111]01=110−111=1110=\int_0^1(1-u)u^9\,du=\int_0^1(u^9-u^{10})du=\Big[\dfrac{u^{10}}{10}-\dfrac{u^{11}}{11}\Big]_0^1=\dfrac{1}{10}-\dfrac{1}{11}=\dfrac{1}{110}.

Answer: (iv) 1/1101/110.

(b) ∫01(2x+3)dx\displaystyle\int_0^1(2x+3)dx as a limit of a sum. With a=0,b=1,h=1na=0,b=1,h=\dfrac1n, f(x)=2x+3f(x)=2x+3:

∫abf(x)dx=lim⁡n→∞h∑r=0n−1f(a+rh)=lim⁡n→∞1n∑r=0n−1(2rn+3)\displaystyle\int_a^bf(x)dx=\lim_{n\to\infty}h\sum_{r=0}^{n-1}f(a+rh)=\lim_{n\to\infty}\frac1n\sum_{r=0}^{n-1}\Big(\frac{2r}{n}+3\Big) …

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