+0  
 
0
327
1
avatar

find the area between the curves y=x^2 and y=1-x^2

Guest Jan 30, 2015

Best Answer 

 #1
avatar+19632 
+10

find the area between the curves y=x^2 and y=1-x^2

I. limits above and below the integral sign:

The Cut of two functions :

$$\\ y_c = x_c^2 = 1-x_c^2 \\
2x_c^2=1 \\
x_c^2=\frac{1}{2} \\
x_c=\pm\sqrt{\frac{1}{2}}\\
x_c=\pm\frac{\sqrt{2}}{2}$$
 

II. The area between is:

 $$\small{\text{
$
\int\limits_{ x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }(1-x^2)-x^2 \ dx
= \int\limits_{x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } 1-2x^2 \ dx
= \int\limits_{x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } dx - 2\int\limits_{x=-1}^{+1}x^2 \ dx
$
}}\\\\
\small{\text{
$
=[x]_{- \frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }-2[\frac{x^3}{3}]_{ -\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }
=[\frac{ \sqrt{2} }{2} -(-\frac{\sqrt{2}}{2} )]-2[\frac{ (\frac{\sqrt{2}}{2})^3}{3}-\frac{(-(\frac{\sqrt{2}}{2}) )^3}{3}]
=\sqrt{2} - 2[\frac{4*\sqrt{2}}{24}-\frac{-4*\sqrt{2}}{24}]
$
}}\\\\
\small{\text{
$
=\sqrt{2} -2[ \frac{2*4 \sqrt{2} }{24}]
=\sqrt{2} - \frac{2}{3} \sqrt{2}
=\frac{1}{3} \sqrt{2}
$
}} \\\\
\small{\text{The area is $ \frac{1}{3} \sqrt{2} $}}$$

heureka  Jan 30, 2015
 #1
avatar+19632 
+10
Best Answer

find the area between the curves y=x^2 and y=1-x^2

I. limits above and below the integral sign:

The Cut of two functions :

$$\\ y_c = x_c^2 = 1-x_c^2 \\
2x_c^2=1 \\
x_c^2=\frac{1}{2} \\
x_c=\pm\sqrt{\frac{1}{2}}\\
x_c=\pm\frac{\sqrt{2}}{2}$$
 

II. The area between is:

 $$\small{\text{
$
\int\limits_{ x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }(1-x^2)-x^2 \ dx
= \int\limits_{x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } 1-2x^2 \ dx
= \int\limits_{x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } dx - 2\int\limits_{x=-1}^{+1}x^2 \ dx
$
}}\\\\
\small{\text{
$
=[x]_{- \frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }-2[\frac{x^3}{3}]_{ -\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }
=[\frac{ \sqrt{2} }{2} -(-\frac{\sqrt{2}}{2} )]-2[\frac{ (\frac{\sqrt{2}}{2})^3}{3}-\frac{(-(\frac{\sqrt{2}}{2}) )^3}{3}]
=\sqrt{2} - 2[\frac{4*\sqrt{2}}{24}-\frac{-4*\sqrt{2}}{24}]
$
}}\\\\
\small{\text{
$
=\sqrt{2} -2[ \frac{2*4 \sqrt{2} }{24}]
=\sqrt{2} - \frac{2}{3} \sqrt{2}
=\frac{1}{3} \sqrt{2}
$
}} \\\\
\small{\text{The area is $ \frac{1}{3} \sqrt{2} $}}$$

heureka  Jan 30, 2015

10 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.