Cool integral property

Evaluate the integral: αβ(xα)(xβ)dx \int_\alpha^\beta (x-\alpha)(x-\beta) dx

SOLUTION αβ(x2(α+β)x+αβ)dx \int_\alpha^\beta (x^2 - (\alpha + \beta)x + \alpha\beta ) dx

(x33(α+β)x22+αβx)αβ \left(\frac{x^3}{3} - \frac{(\alpha + \beta)x^2}{2} + \alpha\beta x\right)\bigg|_{\alpha}^{\beta}

13(β3α3)α+β2(β2α2)+αβ(βα) \frac{1}{3}(\beta^3 - \alpha^3)-\frac{\alpha+\beta}{2}(\beta^2 - \alpha^2) + \alpha\beta(\beta-\alpha)

13(βα)(β2+αβ+α2)α+β2(βα)(β+α)+αβ(βα) \frac{1}{3}(\beta - \alpha)(\beta^2+\alpha\beta+\alpha^2)-\frac{\alpha+\beta}{2}(\beta - \alpha)(\beta + \alpha) + \alpha\beta(\beta-\alpha)

(βα)(13β2+13αβ+13α212α2αβ12β2+αβ) (\beta-\alpha)\left(\frac{1}{3}\beta^2 + \frac{1}{3}\alpha\beta + \frac{1}{3}\alpha^2 - \frac{1}{2}\alpha^2 - \alpha\beta - \frac{1}{2}\beta^2 + \alpha\beta\right)

(βα)(16β2+13αβ16α2) (\beta-\alpha)\left(-\frac{1}{6}\beta^2 +\frac{1}{3}\alpha\beta - \frac{1}{6}\alpha^2\right)

(βα)[(βα)26](\beta-\alpha)\left[\frac{(\beta-\alpha)^2}{-6}\right]

(βα)36\underline{\underline{-\frac{(\beta-\alpha)^3}{6}}}

αβ(xα)(xβ)dx=(βα)36\boxed{\int_\alpha^\beta (x-\alpha)(x-\beta) dx = -\frac{(\beta-\alpha)^3}{6}}

Note by JohnDonnie Celestre
4 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Couldn't we substitute x - average(alpha,beta) = u ?

That would have significantly reduced the computational time. Please think over it

Soutrik Bandyopadhyay - 4 years, 9 months ago

Log in to reply

yeah thanks.

JohnDonnie Celestre - 4 years, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...