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Evaluate the integral by using area formulas from geometry?

integral from 1/2 to 1 of √(1-(x^2))

i can’t figure this one out.
i would like to avoid plotting but yes this is the way i am asked to do this. otherwise i know how to do it

You cannot integrate by power rule. You would have to use trig substitution.

However, what the teacher wants is for you to graph it.
This is the top half of the circle, x^2+y^2=1
This circle has center at the origin and radius =1

Then find the area of the strip from x= 1/2 to x=1

Draw a radius from (0,0) to (1/2, sqr3/2)
Between this radius and the x axis is a sector, with angle=60 degrees (special triangle formula)
Area(sector) = (m/360)(pi* r^2)= pi/6

Now the region from 0 to 1/2 is a right triangle with area = (1/2)(1/2)(sqr(3)/2= sqr(3)/8

So area from 0 to 1/2 is [pi/6 - sqr(3)/8]

You get the same answer if you integrate by trig subst.

Hoping this helps!

This entry was posted on Sunday, February 12th, 2012 at 9:13 pm and is filed under area formulas. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Evaluate the integral by using area formulas from geometry?”

  1. ∫ 1/2 to 1 √(1-x^2) dx

    Root is to the power of a 1/2

    ∫ 1/2 to 1 (1-x^2)^1/2 dx
    Using the formula
    ∫(ax+b)^n = [1/n(ax+b)^n+1]/a(n+1)
    we get
    [-1(1-x^2)^3/2]/(3/2(-1) from 1/2 to 1
    [2(1-x^2)^3/2]/3 from 1/2 to 1

    Substitute and solve from there.

    Unless you are given a graph you can’t really solve geometrically unless you intend on plotting.
    References :

  2. You cannot integrate by power rule. You would have to use trig substitution.

    However, what the teacher wants is for you to graph it.
    This is the top half of the circle, x^2+y^2=1
    This circle has center at the origin and radius =1

    Then find the area of the strip from x= 1/2 to x=1

    Draw a radius from (0,0) to (1/2, sqr3/2)
    Between this radius and the x axis is a sector, with angle=60 degrees (special triangle formula)
    Area(sector) = (m/360)(pi* r^2)= pi/6

    Now the region from 0 to 1/2 is a right triangle with area = (1/2)(1/2)(sqr(3)/2= sqr(3)/8

    So area from 0 to 1/2 is [pi/6 - sqr(3)/8]

    You get the same answer if you integrate by trig subst.

    Hoping this helps!
    References :

  3. Zonic the hedgehog on February 13th, 2012 at 3:37 am

    1
    ∫ √(1 – x²) dx
    1/2

    Let :
    x = sinu ( x ∈ [-π/2 ; π/2 ] )
    dx = cosu.du

    Change :
    x __| __ 1/2 ___ 1
    u __| __ π/6 ___ π/2

    . .π/2
    = ∫ . . √(1 – sin²u).cosu.du
    . .π/6

    . .π/2
    = ∫ . . cosu.cosu.du
    . .π/6

    . . π/2
    = ∫ . . cos²u .du
    . . π/6

    . .π/2 . 1 + cos2u
    = ∫ . . . ▬▬▬▬▬ du
    . .π/6 . . . . 2

    . .π/2 . 1 . . . . . . 1 .π/2
    = ∫ . . ▬▬ du + ▬▬∫ . cos2u du
    . .π/6 . 2. . . . . . .2 .π/6

    . . .1 . . . .π/2 . . . . 1 . . . . . . . . π/2
    = ▬▬(u) | . . . + ▬▬▬[ sin(2u)] |
    . . .2 .. . . π/6 . . . . 4 . . . . . . . . π/6

    . . .1 . . . . . . . . . . . . .1
    = ▬▬( π/2 – π/6 ) + ▬▬[ sin(2π/2) - sin(2π/6) ]
    . . .2 . . . . . . . . . . . . .4

    . . .π . . . √3
    = ▬▬ – ▬▬
    . . .6 . . . .8
    References :
    I’m a Vietnamese student grade 12

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