Question #2e036

Question

Question #2e036

2 Answers

Impact of this question

1236 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-21T21:09:22

Shown by considering geometry...

Explanation:

I wanted to provide a simple way of evaluating this integral

The first thing we see is that the function is $y = \sqrt{1 - x^{2}}$ $y = sqrt(1-x^2)$

$\Rightarrow y^{2} = 1 - x^{2}$ $=> y^2 = 1 - x^2$

$\Rightarrow x^{2} + y^{2} = 1$ $=> x^2+y^2 = 1$

This is just a circle of radius 1, and centre $(0, 0)$ $(0,0)$

This is just this shaded region...

When evaluating from $x = 0$ $x = 0$ and $x = 1$ $x=1$

Desmos

What is just a quarter of the area of a circle of radius 1

Full area of the circle = $= π r^{2} = π \cdot 1^{2} = π$ $= pir^2 =pi * 1^2 = pi$

Hence quarter area = $\frac{1}{4} \cdot π$ $1/4 * pi$

$= \frac{π}{4}$ $color(red)(= pi/4$

Answer 2 · 2018-01-22T06:46:01

Conventoinal means of substitution...

Explanation:

The conventional way of solving this integral:

We can do a trigonometric substitution...

Let $x = sin θ$ $x = sintheta$

$d x = cos θ d θ$ $color(green)(dx = costheta d theta$

and we know ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$

$\Rightarrow cos θ = \sqrt{1 - {sin}^{2} θ}$ $=> costheta = sqrt(1-sin^2 theta )$

$\Rightarrow cos θ = \sqrt{1 - x^{2}}$ $=> color(blue)(costheta = sqrt(1-x^2)$

Hence our integral becomes:

$\int cos θ cos θ d θ$ $int color(blue)(costheta) color(green)(costheta d theta$

Changing the limits...

$θ = {sin}^{- 1} x$ $theta = sin^(-1) x$

$x = 1 \Rightarrow θ = {sin}^{- 1} (1) = \frac{π}{2}$ $x = 1 => theta = sin^(-1)(1) = pi/2$

$x = 0 \Rightarrow θ = {sin}^{- 1} (0) = 0$ $x = 0 => theta = sin^(-1)(0) = 0$

$\Rightarrow \int_{0}^{\frac{π}{2}} {cos}^{2} θ d θ$ $=> int_0 ^(pi/2) cos^2 theta d theta$

Use $cos 2 θ = {cos}^{2} θ - {sin}^{2} θ$ $cos2theta = cos^2 theta - sin^2 theta$

$\Rightarrow cos 2 θ = 2 {cos}^{2} θ - 1$ $=> cos2theta = 2cos^2 theta -1$

${cos}^{2} θ = \frac{1}{2} (cos 2 θ + 1)$ $cos^2 theta = 1/2( cos2theta + 1 )$

$\Rightarrow \frac{1}{2} \int_{0}^{\frac{π}{2}} cos 2 θ + 1 d θ$ $=>1/2 int_0 ^(pi/2) cos2theta + 1 d theta$

$\Rightarrow \frac{1}{2} {[\frac{1}{2} sin 2 θ + θ]}_{0}^{\frac{π}{2}}$ $=> 1/2[ 1/2 sin2theta + theta ]_0 ^(pi/2)$

$\Rightarrow \frac{1}{2} [{0 + \frac{π}{2}} - {0 + 0}]$ $=> 1/2[ {0 + pi/2}-{0 + 0 } ]$

$= \frac{π}{4}$ $color(blue)( = pi/4 )$

Science

Math

Humanities

... and beyond

Question #2e036

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question