Question Number 101319 by mhmd last updated on 01/Jul/20
$${find}\:{the}\:{area}\:{bounded}\:{the}\:{parabola}\: \\ $$$${y}=\mathrm{4}{x}^{\mathrm{2}} \:\:\:{and}\:\:{y}=\mathrm{8}−\mathrm{4}{x}^{\mathrm{2}} \:\:?\:\:{by}\:{using}\:{intigral}? \\ $$$${help}\:{me} \\ $$
Commented by smridha last updated on 01/Jul/20
![the area under the curves(here parabola) A=2[∫_0 ^1 (8−4x^2 )dx−∫_0 ^1 4x^2 dx] =2[8x−(8/3)x^3 ]_0 ^1 =2[(8/1)−(8/3)]=((32)/3).](https://www.tinkutara.com/question/Q101321.png)
$$\boldsymbol{{the}}\:\boldsymbol{{area}}\:\boldsymbol{{under}}\:\boldsymbol{{the}}\:\boldsymbol{{curves}}\left(\boldsymbol{{here}}\:\boldsymbol{{parabola}}\right) \\ $$$$\boldsymbol{{A}}=\mathrm{2}\left[\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{8}−\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \right)\boldsymbol{{dx}}−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{dx}}\right] \\ $$$$\:\:\:\:=\mathrm{2}\left[\mathrm{8}\boldsymbol{{x}}−\frac{\mathrm{8}}{\mathrm{3}}\boldsymbol{{x}}^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2}\left[\frac{\mathrm{8}}{\mathrm{1}}−\frac{\mathrm{8}}{\mathrm{3}}\right]=\frac{\mathrm{32}}{\mathrm{3}}. \\ $$
Commented by mhmd last updated on 01/Jul/20
$${very}\:{thank}\:{sir} \\ $$
Commented by smridha last updated on 01/Jul/20
welcome...I think you can draw the picture..
Commented by mhmd last updated on 01/Jul/20
$${yes}\:{sir} \\ $$
Answered by bobhans last updated on 01/Jul/20
![intercept : 4x^2 = 8−4x^2 ; 8x^2 = 8 ; x = ± 1 test x = (1/2) → { ((y=4×(1/4)=1)),((y=8−4×(1/4)= 7)) :} 8−4x^2 ≥ 4x^2 , so the area = 2∫_0 ^1 (8−8x^2 )dx = 2[(8x−(8/3)x^3 )]_0 ^1 = 2(8−(8/3)) = 2×((16)/3) = ((32)/3) ★](https://www.tinkutara.com/question/Q101324.png)
$$\mathrm{intercept}\::\:\mathrm{4x}^{\mathrm{2}} \:=\:\mathrm{8}−\mathrm{4x}^{\mathrm{2}} \:;\:\mathrm{8x}^{\mathrm{2}} \:=\:\mathrm{8}\:;\:\mathrm{x}\:=\:\pm\:\mathrm{1} \\ $$$$\mathrm{test}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\rightarrow\begin{cases}{\mathrm{y}=\mathrm{4}×\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{1}}\\{\mathrm{y}=\mathrm{8}−\mathrm{4}×\frac{\mathrm{1}}{\mathrm{4}}=\:\mathrm{7}}\end{cases} \\ $$$$\mathrm{8}−\mathrm{4x}^{\mathrm{2}} \:\geqslant\:\mathrm{4x}^{\mathrm{2}} \:,\:\mathrm{so}\:\mathrm{the}\:\mathrm{area}\:=\: \\ $$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{8}−\mathrm{8x}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\mathrm{2}\left[\left(\mathrm{8x}−\frac{\mathrm{8}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\mathrm{2}\left(\mathrm{8}−\frac{\mathrm{8}}{\mathrm{3}}\right)\:=\:\mathrm{2}×\frac{\mathrm{16}}{\mathrm{3}}\:=\:\frac{\mathrm{32}}{\mathrm{3}}\:\bigstar \\ $$$$ \\ $$$$ \\ $$
Commented by bramlex last updated on 02/Jul/20
$$ \Box\spadesuit\bigstar \\ $$
Commented by mhmd last updated on 02/Jul/20
$${wow}\:!\:{very}\:{nice}\:{sir}\:{thank}\:{you} \\ $$