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This is one of our many quizzes on Reverse power rule. Click on this to access more such competitive quizzes on Calculus.

Reverse power rule Quiz 2

mathematics

 Which power of $x$ in the reverse power rule does not give an antiderivative..

 Reverse Power Rule does not give an antiderivative when its integration cannot be defined.

  $\text{Lets us integrate any power of “x” } (\text{negative or a decimal})\text{ other than “-1”.} \\ \text{The Reverse power rule is given by } \\  \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}. \\ \text{Here, if we put value of n=-1 then,} \\ \Rightarrow \int{{{x}^{-1}}dx=\dfrac{{{x}^{-1+1}}}{-1+1}+C} \\ \Rightarrow \dfrac{{{x}^{0}}}{0}=\text{ Not defined}$

 What would be the value  of  $\int{\left( {{x}^{3}}+\dfrac{3}{2}{{x}^{2}}-2x \right)}dx$ using reverse power rule.

 Integrate using reverse power rule , given by,   $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$.

  $\text{Integrating }\int{\left( {{x}^{3}}+\dfrac{3}{2}{{x}^{2}}-2x \right)}dx\text{ we get, } \\ \int{\left( {{x}^{3}}+\dfrac{3}{2}{{x}^{2}}-2x \right)}dx=\int{{{x}^{3}}}dx+\int{\dfrac{3}{2}{{x}^{2}}}dx-\int{2x}dx \\ \Rightarrow \dfrac{{{x}^{3+1}}}{3+1}+\dfrac{3}{2}\times \dfrac{{{x}^{2+1}}}{2+1}-2\times \dfrac{{{x}^{1+1}}}{1+1} \\ \Rightarrow \dfrac{{{x}^{4}}}{4}+\dfrac{3}{2}\times \dfrac{{{x}^{3}}}{3}-2\times \dfrac{{{x}^{2}}}{2} \\ \Rightarrow \dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{3}}}{2}-{{x}^{2}}+C $

 Integrate $\dfrac{2\sqrt[3]{x}}{3}+2{{x}^{5}}+5$  using reverse power rule.

  $\text{Using the Reverse Power rule,} \\ \int{\dfrac{2\sqrt[3]{x}}{3}+2{{x}^{5}}+5\text{ dx}=\int{\dfrac{2}{3}{{x}^{\frac{1}{3}}}+}2{{x}^{5}}+5\text{ dx}} \\ \text{                              =}\dfrac{2}{3}\int{{{x}^{\frac{1}{3}}}}\text{ dx +}2\int{{{x}^{5}}\text{ dx +}\int{5\text{ dx}}} \\ \text{                             }=\dfrac{2}{3}\times \dfrac{{{x}^{\frac{1}{3}+1}}}{^{\frac{1}{3}+1}}+2\times \dfrac{{{x}^{5+1}}}{5+1}+5x+C \\ \text{                            =}\dfrac{2}{3}\times \dfrac{{{x}^{\frac{4}{3}}}}{^{\frac{4}{3}}}+2\times \dfrac{{{x}^{6}}}{6}+5x+C \\ \text{                              =}\dfrac{2}{3}\times \dfrac{3}{4}{{x}^{\frac{4}{3}}}+\dfrac{{{x}^{6}}}{3}+5x+C \\ \text{                             =}\dfrac{{{x}^{\frac{4}{3}}}}{2}+\dfrac{{{x}^{6}}}{3}+5x+C \\ \therefore \int{\dfrac{2\sqrt[3]{x}}{3}+2{{x}^{5}}+5\text{ dx}=}\dfrac{{{x}^{\frac{4}{3}}}}{2}+\dfrac{{{x}^{6}}}{3}+5x+C$

 Find the antiderivative of $f(x)=2x$ using reverse power rule and from that determine how the tangents of this family of integrals are related to each other?

 If a family of parabolas intersect by  a line $x=a$, then the $\dfrac{dy}{dx}$ at these points equals to $2a$.The tangents to the curves at these points are parallel.

  $\text{According to reverse power rule,  } \\ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C} \\ \text{The anti derivative of  }f(x)=2x\text{ will be,} \\ \int{2{{x}^{1}}dx=\dfrac{2{{x}^{1+1}}}{1+1}=\dfrac{2{{x}^{2}}}{2}={{x}^{2}}}+C \\ y={{x}^{2}}+C,  \\ \text{ As it can be seen from the image below,} \\ \text{For  } C=1,\text{ we obtain }y={{x}^{2}}+1 \\ C=2,\text{ we obtain }y={{x}^{2}}+2 \\ C=-1,\text{ we obtain }y={{x}^{2}}-1 \\ C=-2,\text{ we obtain }y={{x}^{2}}-2 \\ $ They all form parabolas. Let the intersection of all these parabola be at x=1. From the figure, it can be said that the tangents to the curve are parallel to each other. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/badfcab0-05ff-45da-86b5-f59efde90df86461399884152827216.png"/>

 Integrate  $\int{\left( \dfrac{{{x}^{3}}-{{x}^{2}}+x{{e}^{x}}-4\sqrt{x}+8}{2x} \right)dx}$  using reverse power rule.

  $\text{We have, }\int{\left( \dfrac{{{x}^{3}}-{{x}^{2}}+x{{e}^{x}}-4\sqrt{x}+8}{2x} \right)dx} \\ \Rightarrow \dfrac{1}{2}\left[\int{{{x}^{2}}dx-\int{xdx+\int{{{e}^{x}}dx-\int{\dfrac{4}{\sqrt{x}}dx+8\int{\dfrac{1}{x}dx}}}}} \right] \\ \text{Using reverse power rule,} \\ \Rightarrow \dfrac{1}{2}\left[ \dfrac{{{x}^{2+1}}}{2+1}-\dfrac{{{x}^{1+1}}}{1+1}+{{e}^{x}}-4\left( \dfrac{{{x}^{-\dfrac{1}{2}+1}}}{^{-\dfrac{1}{2}+1}} \right)+8\ln \left| x \right| \right] \\ \Rightarrow \dfrac{1}{2}\left[ \dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+{{e}^{x}}-4\left( \dfrac{\sqrt{x}}{\dfrac{1}{2}} \right)+8\ln \left| x \right| \right] \\ \Rightarrow \dfrac{1}{2}\left( \dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+{{e}^{x}}-8\sqrt{x}+8\ln \left| x \right| \right) \\ \Rightarrow \dfrac{{{x}^{3}}}{6}-\dfrac{{{x}^{2}}}{4}+\dfrac{{{e}^{x}}}{2}-4\left( \sqrt{x}-\ln \left| x \right| \right)$