Need help proving!

\[I=\displaystyle \int^{b}_{a}\dfrac{x^{n-1}\{(n-2)x^2+(n-1)(a+b)x+nab\}}{(x+a)^2(x+b)^2}dx\]

Prove that \(I=\dfrac{b^{n-1}-a^{n-1}}{2(a+b)}\).

Note by Akshat Sharda
1 year, 1 month ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Observe that \((x+b)^2 - (x+a)^2 = (b-a)(2x+a+b)\). Now

\[\begin{align} I &= \int_a^b \dfrac{x^{n-1} \left\{ (n-2)x^2 + (n-1)(a+b)x + nab \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{b-a}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (n-2)x^2 + (n-1)(a+b)x + nab \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{b-a}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ n(x^2 + (a+b)x + ab) - x(2x+a+b) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (b-a)n(x+a)(x+b) - x(b-a)(2x+a+b) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (b-a)n(x+a)(x+b) - x((x+b)^2 - (x+a)^2) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \left[ \dfrac{n(b-a)x^{n-1}}{(x+a)(x+b)} - x^n \left\{ \dfrac{1}{(x+a)^2} - \dfrac{1}{(x+b)^2} \right\} \right] \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \left[ nx^{n-1} \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} - x^n \left\{ \dfrac{1}{(x+a)^2} - \dfrac{1}{(x+b)^2} \right\} \right] \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \mathrm{d} \left( x^n \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} \right) \\ &= \dfrac{1}{b-a} \cdot \left( x^n \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} \right) {\huge |}_a^b \\ &= \dfrac{x^n}{(x+a)(x+b)} {\huge |}_a^b \\ &= \boxed{\dfrac{b^{n-1} - a^{n-1}}{2(a+b)}} \end{align}\]

Tapas Mazumdar - 1 year, 1 month ago

Log in to reply

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.

Tapas Mazumdar - 1 year ago

Log in to reply

Really good solution! (+0!)

Harsh Shrivastava - 1 year, 1 month ago

Log in to reply

Haha! :P

Tapas Mazumdar - 1 year, 1 month ago

Log in to reply

Great solution! (+1)

Akshat Sharda - 1 year, 1 month ago

Log in to reply

Thank you. :)

Tapas Mazumdar - 1 year, 1 month ago

Log in to reply

@Tapas Mazumdar Are you also preparing for JEE?

Akshat Sharda - 1 year, 1 month ago

Log in to reply

@Akshat Sharda Yes. I'm studying in FIITJEE at Raipur, CG.

Tapas Mazumdar - 1 year, 1 month ago

Log in to reply

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]

Rahul Sharma - 1 year ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...