I=∫abxn−1{(n−2)x2+(n−1)(a+b)x+nab}(x+a)2(x+b)2dxI=\displaystyle \int^{b}_{a}\dfrac{x^{n-1}\{(n-2)x^2+(n-1)(a+b)x+nab\}}{(x+a)^2(x+b)^2}dxI=∫ab(x+a)2(x+b)2xn−1{(n−2)x2+(n−1)(a+b)x+nab}dx
Prove that I=bn−1−an−12(a+b)I=\dfrac{b^{n-1}-a^{n-1}}{2(a+b)}I=2(a+b)bn−1−an−1.
Note by Akshat Sharda 3 years, 9 months ago
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Observe that (x+b)2−(x+a)2=(b−a)(2x+a+b)(x+b)^2 - (x+a)^2 = (b-a)(2x+a+b)(x+b)2−(x+a)2=(b−a)(2x+a+b). Now
I=∫abxn−1{(n−2)x2+(n−1)(a+b)x+nab}(x+a)2(x+b)2 dx=b−ab−a⋅∫abxn−1{(n−2)x2+(n−1)(a+b)x+nab}(x+a)2(x+b)2 dx=b−ab−a⋅∫abxn−1{n(x2+(a+b)x+ab)−x(2x+a+b)}(x+a)2(x+b)2 dx=1b−a⋅∫abxn−1{(b−a)n(x+a)(x+b)−x(b−a)(2x+a+b)}(x+a)2(x+b)2 dx=1b−a⋅∫abxn−1{(b−a)n(x+a)(x+b)−x((x+b)2−(x+a)2)}(x+a)2(x+b)2 dx=1b−a⋅∫ab[n(b−a)xn−1(x+a)(x+b)−xn{1(x+a)2−1(x+b)2}] dx=1b−a⋅∫ab[nxn−1{1x+a−1x+b}−xn{1(x+a)2−1(x+b)2}] dx=1b−a⋅∫abd(xn{1x+a−1x+b})=1b−a⋅(xn{1x+a−1x+b})∣ab=xn(x+a)(x+b)∣ab=bn−1−an−12(a+b)\begin{aligned} I &= \int_a^b \dfrac{x^{n-1} \left\{ (n-2)x^2 + (n-1)(a+b)x + nab \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{b-a}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (n-2)x^2 + (n-1)(a+b)x + nab \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{b-a}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ n(x^2 + (a+b)x + ab) - x(2x+a+b) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (b-a)n(x+a)(x+b) - x(b-a)(2x+a+b) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (b-a)n(x+a)(x+b) - x((x+b)^2 - (x+a)^2) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \left[ \dfrac{n(b-a)x^{n-1}}{(x+a)(x+b)} - x^n \left\{ \dfrac{1}{(x+a)^2} - \dfrac{1}{(x+b)^2} \right\} \right] \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \left[ nx^{n-1} \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} - x^n \left\{ \dfrac{1}{(x+a)^2} - \dfrac{1}{(x+b)^2} \right\} \right] \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \mathrm{d} \left( x^n \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} \right) \\ &= \dfrac{1}{b-a} \cdot \left( x^n \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} \right) {\huge |}_a^b \\ &= \dfrac{x^n}{(x+a)(x+b)} {\huge |}_a^b \\ &= \boxed{\dfrac{b^{n-1} - a^{n-1}}{2(a+b)}} \end{aligned}I=∫ab(x+a)2(x+b)2xn−1{(n−2)x2+(n−1)(a+b)x+nab} dx=b−ab−a⋅∫ab(x+a)2(x+b)2xn−1{(n−2)x2+(n−1)(a+b)x+nab} dx=b−ab−a⋅∫ab(x+a)2(x+b)2xn−1{n(x2+(a+b)x+ab)−x(2x+a+b)} dx=b−a1⋅∫ab(x+a)2(x+b)2xn−1{(b−a)n(x+a)(x+b)−x(b−a)(2x+a+b)} dx=b−a1⋅∫ab(x+a)2(x+b)2xn−1{(b−a)n(x+a)(x+b)−x((x+b)2−(x+a)2)} dx=b−a1⋅∫ab[(x+a)(x+b)n(b−a)xn−1−xn{(x+a)21−(x+b)21}] dx=b−a1⋅∫ab[nxn−1{x+a1−x+b1}−xn{(x+a)21−(x+b)21}] dx=b−a1⋅∫abd(xn{x+a1−x+b1})=b−a1⋅(xn{x+a1−x+b1})∣ab=(x+a)(x+b)xn∣ab=2(a+b)bn−1−an−1
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Great solution! (+1)
Thank you. :)
@Tapas Mazumdar – Are you also preparing for JEE?
@Akshat Sharda – Yes. I'm studying in FIITJEE at Raipur, CG.
Really good solution! (+0!)
Haha! :P
WHAT THE HECK IS THAT! I know its calculus but that is a very long equation.
Well for a math lover like me, I like to deal with beautiful long expressions that turn finally to a small answer.
Hello, Some of the important books for preparing JEE Maths are ML KHANNA IA MARRON for basic concepts SL Loney for Coordinate geometry SL Loney for Trigonometry Hall knight for Higher Algebra You must also check out for [url="https://scoop.eduncle.com/jee-advanced-exam-date-notification"] </a>JEE Advanced Date 2018 [/url]
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Observe that (x+b)2−(x+a)2=(b−a)(2x+a+b). Now
I=∫ab(x+a)2(x+b)2xn−1{(n−2)x2+(n−1)(a+b)x+nab} dx=b−ab−a⋅∫ab(x+a)2(x+b)2xn−1{(n−2)x2+(n−1)(a+b)x+nab} dx=b−ab−a⋅∫ab(x+a)2(x+b)2xn−1{n(x2+(a+b)x+ab)−x(2x+a+b)} dx=b−a1⋅∫ab(x+a)2(x+b)2xn−1{(b−a)n(x+a)(x+b)−x(b−a)(2x+a+b)} dx=b−a1⋅∫ab(x+a)2(x+b)2xn−1{(b−a)n(x+a)(x+b)−x((x+b)2−(x+a)2)} dx=b−a1⋅∫ab[(x+a)(x+b)n(b−a)xn−1−xn{(x+a)21−(x+b)21}] dx=b−a1⋅∫ab[nxn−1{x+a1−x+b1}−xn{(x+a)21−(x+b)21}] dx=b−a1⋅∫abd(xn{x+a1−x+b1})=b−a1⋅(xn{x+a1−x+b1})∣ab=(x+a)(x+b)xn∣ab=2(a+b)bn−1−an−1
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Great solution! (+1)
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Thank you. :)
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Are you also preparing for JEE?
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Yes. I'm studying in FIITJEE at Raipur, CG.
Really good solution! (+0!)
Log in to reply
Haha! :P
WHAT THE HECK IS THAT! I know its calculus but that is a very long equation.
Log in to reply
Well for a math lover like me, I like to deal with beautiful long expressions that turn finally to a small answer.
Hello, Some of the important books for preparing JEE Maths are ML KHANNA IA MARRON for basic concepts SL Loney for Coordinate geometry
SL Loney for Trigonometry Hall knight for Higher Algebra You must also check out for [url="https://scoop.eduncle.com/jee-advanced-exam-date-notification"] </a>JEE Advanced Date 2018 [/url]