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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

$$\boldsymbol{{the}}\:\boldsymbol{{area}}\:\boldsymbol{{under}}\:\boldsymbol{{the}}\:\boldsymbol{{curves}}\left(\boldsymbol{{here}}\:\boldsymbol{{parabola}}\right) \\$$$$\boldsymbol{{A}}=\mathrm{2}\left[\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{8}−\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \right)\boldsymbol{{dx}}−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{dx}}\right] \\$$$$\:\:\:\:=\mathrm{2}\left[\mathrm{8}\boldsymbol{{x}}−\frac{\mathrm{8}}{\mathrm{3}}\boldsymbol{{x}}^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2}\left[\frac{\mathrm{8}}{\mathrm{1}}−\frac{\mathrm{8}}{\mathrm{3}}\right]=\frac{\mathrm{32}}{\mathrm{3}}. \\$$

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

$$\mathrm{intercept}\::\:\mathrm{4x}^{\mathrm{2}} \:=\:\mathrm{8}−\mathrm{4x}^{\mathrm{2}} \:;\:\mathrm{8x}^{\mathrm{2}} \:=\:\mathrm{8}\:;\:\mathrm{x}\:=\:\pm\:\mathrm{1} \\$$$$\mathrm{test}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\rightarrow\begin{cases}{\mathrm{y}=\mathrm{4}×\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{1}}\\{\mathrm{y}=\mathrm{8}−\mathrm{4}×\frac{\mathrm{1}}{\mathrm{4}}=\:\mathrm{7}}\end{cases} \\$$$$\mathrm{8}−\mathrm{4x}^{\mathrm{2}} \:\geqslant\:\mathrm{4x}^{\mathrm{2}} \:,\:\mathrm{so}\:\mathrm{the}\:\mathrm{area}\:=\: \\$$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{8}−\mathrm{8x}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\mathrm{2}\left[\left(\mathrm{8x}−\frac{\mathrm{8}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$=\:\mathrm{2}\left(\mathrm{8}−\frac{\mathrm{8}}{\mathrm{3}}\right)\:=\:\mathrm{2}×\frac{\mathrm{16}}{\mathrm{3}}\:=\:\frac{\mathrm{32}}{\mathrm{3}}\:\bigstar \\$$$$\\$$$$\\$$

Commented by bramlex last updated on 02/Jul/20

$$\Box\spadesuit\bigstar \\$$

Commented by mhmd last updated on 02/Jul/20

$${wow}\:!\:{very}\:{nice}\:{sir}\:{thank}\:{you} \\$$

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