Chapter 8: Applications of Integrals

Q1Written answer

Find the area under the given curves and given lines:

(i) \[\text{y=}{{\text{x}}^{\text{2}}}\text{,x=1,x=2}\] and \[\text{x-axis}\]

(ii) \[\text{y=}{{\text{x}}^{4}}\text{,x=1,x=5}\] and \[\text{x-axis}\]

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\[\text{AreaADCBA=}\int_{\text{1}}^{\text{2}}{\text{y}}\text{dx}\]

substitute $y={{x}^{2}}$

\[\text{=}\int_{\text{1}}^{\text{2}}{{{\text{x}}^{\text{2}}}}\text{dx}\]

Substituting the limits,

\[\text{=}\frac{\text{8}}{\text{3}}\text{-}\frac{\text{1}}{\text{3}}\]

\[\text{=}\frac{\text{7}}{\text{3}}\text{units}\]

Ans:

\[\text{AreaofADCBA=}\int_{\text{1}}^{\text{5}}{{{\text{x}}^{\text{4}}}}\text{dx}\]

Integrating using the power rule,

\[\text{=}\left[ \frac{{{\text{x}}^{\text{5}}}}{\text{5}} \right]_{\text{1}}^{\text{5}}\]

Substituting the limits,

\[\text{=}\frac{{{\text{(5)}}^{\text{5}}}}{\text{5}}\text{-}\frac{\text{1}}{\text{5}}\]

Simplifying,

\[\text{=(5}{{\text{)}}^{\text{4}}}\text{-}\frac{\text{1}}{\text{5}}\]

\[\text{=625-}\frac{\text{1}}{\text{5}}\]

\[\text{=624}\text{.8 units}\]

Q2Written answer

Sketch the graph of \[\text{y=}\left| \text{x+3} \right|\] and evaluate \[\int_{-6}^{0}{\left| \text{x+3} \right|}\text{dx}\].

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\[\text{x}\]	\[\text{-6}\]	\[\text{-5}\]	\[\text{-4}\]	\[\text{-3}\]	\[\text{-2}\]	\[\text{-1}\]	\[\text{0}\]
\[\text{y}\]	\[\text{3}\]	\[\text{2}\]	\[\text{1}\]	\[\text{0}\]	\[\text{1}\]	\[\text{2}\]	\[\text{3}\]

\[\left( \text{x+3} \right)\le 0\] for \[\text{-6}\le \text{x}\le \text{-3}\]

\[\left( \text{x+3} \right)\ge \text{0}\] for \[\text{-3}\le \text{x}\le 0\]

Therefore,

\[\int_{\text{-6}}^{\text{0}}{\text{ }\!\!|\!\!\text{ }}\text{(x+3) }\!\!|\!\!\text{ dx=-}\int_{\text{-6}}^{\text{-3}}{\text{(x+3)}}\text{dx+}\int_{\text{-3}}^{\text{0}}{\text{(x+3)}}\text{dx}\]

Integrating using the power rule

\[\text{= -}\left[ \frac{{{\text{x}}^{\text{2}}}}{\text{2}}\text{+3x} \right]_{\text{-6}}^{\text{-3}}\text{+}\left[ \frac{{{\text{x}}^{\text{2}}}}{\text{2}}\text{+3x} \right]_{\text{-3}}^{\text{0}}\]

Substituting the limits,

\[\text{=-}\left[ \left( \frac{{{\text{(-3)}}^{\text{2}}}}{\text{2}}\text{+3(-3)} \right)\text{-}\left( \frac{{{\text{(-6)}}^{\text{2}}}}{\text{2}}\text{+3(-6)} \right) \right]\text{+}\left[ \text{0-}\left( \frac{{{\text{(-3)}}^{\text{2}}}}{\text{2}}\text{+3(-3)} \right) \right]\]

Simplifying,

\[\text{= -}\left[ \text{-}\frac{\text{9}}{\text{2}} \right]\text{-}\left[ \text{-}\frac{\text{9}}{\text{2}} \right]\]

\[\text{=9}\]

Q3Written answer

Find the area bounded by the curve \[\text{y=sinx}\] between \[\text{x=0}\] and \[\text{x=2 }\!\!\pi\!\!\text{ }\].

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Therefore, \[\text{area = Area OABO+ Area BCDB}\]

Area Bounded by Curve y=sinx

\[\text{=}\int_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}{\text{sin}}\text{xdx+}\left| \int_{\text{ }\!\!\pi\!\!\text{ }}^{\text{2 }\!\!\pi\!\!\text{ }}{\text{sin}}\text{xdx} \right|\]

\[\text{= }\!\![\!\!\text{ -cosx }\!\!]\!\!\text{ }_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}\text{+}\left| \text{ }\!\![\!\!\text{ -cosx }\!\!]\!\!\text{ }_{\text{ }\!\!\pi\!\!\text{ }}^{\text{2 }\!\!\pi\!\!\text{ }} \right|\]

Substituting the limits,

\[\text{= }\!\![\!\!\text{ -cos }\!\!\pi\!\!\text{ +cos0 }\!\!]\!\!\text{ + }\!\!|\!\!\text{ -cos2 }\!\!\pi\!\!\text{ +cos }\!\!\pi\!\!\text{ }\!\!|\!\!\text{ }\]

Simplifying,

\[\text{=1+1+ }\!\!|\!\!\text{ (-1-1) }\!\!|\!\!\text{ }\]

\[\text{=2+ }\!\!|\!\!\text{ -2 }\!\!|\!\!\text{ }\]

\[\text{=2+2}\]

\[\text{=4 units}\]

Q4Multiple choice

Area bounded by the curve \[\text{y=}{{\text{x}}^{3}}\], the \[\text{x-axis}\] and the ordinates \[\text{x = -2}\] and \[\text{x = 1}\] is

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As shown in the diagram, the required area is:

\[\text{Required area =}\int_{\text{-2}}^{\text{1}}{\text{y}}\text{dx}\]

\[\text{=}\int_{\text{-2}}^{\text{1}}{{{\text{x}}^{\text{3}}}}\text{dx}\]

Integrating using the power rule

\[\text{=}\left[ \frac{{{\text{x}}^{\text{4}}}}{\text{4}} \right]_{\text{-2}}^{\text{1}}\]

Substituting the limits,

\[\text{=}\left[ \frac{\text{1}}{\text{4}}\text{-}\frac{{{\text{(-2)}}^{\text{4}}}}{\text{4}} \right]\]

Simplifying,

\[\text{=}\left( \frac{\text{1}}{\text{4}}\text{-4} \right)\]

\[\text{= -}\frac{\text{15}}{\text{4}}\text{ units}\]

So, the correct answer is \[\text{ -}\frac{\text{15}}{\text{4}}\text{ units}\] option B.

Q5Multiple choice

The area bounded by the curve \[\text{y=x}\left| \text{x} \right|\text{,x-axis}\] and the ordinate \[\text{x = 1}\] and \[\text{x = -1}\] is given by (Hint \[\text{y = }{{\text{x}}^{2}}\] if \[x>0\] and \[\text{y = -}{{\text{x}}^{2}}\] if \[x<0\])

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\[\text{Required area=}\int_{\text{-1}}^{\text{1}}{\text{y}}\text{dx}\]

\[\text{=}\int_{\text{-1}}^{\text{1}}{\text{x}}\text{ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ dx}\]

\[\text{= -}\int_{\text{-1}}^{\text{0}}{{{\text{x}}^{\text{2}}}}\text{dx+}\int_{\text{0}}^{\text{1}}{{{\text{x}}^{\text{2}}}}\text{dx}\]

Integrating using the power rule

\[\text{= -}\left[ \frac{{{\text{x}}^{\text{3}}}}{\text{3}} \right]_{\text{-1}}^{\text{0}}\text{+}\left[ \frac{{{\text{x}}^{\text{3}}}}{\text{3}} \right]_{\text{0}}^{\text{1}}\]

Substituting the limits,

\[\text{= -}\left( \text{-}\frac{\text{1}}{\text{3}} \right)\text{+}\frac{\text{1}}{\text{3}}\]

\[\text{=}\frac{\text{2}}{\text{3}}\text{units}\]

So, the correct answer is \[\frac{\text{2}}{\text{3}}\text{units}\] option C.

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