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This is the second of our many quizzes on Definite integrals. Click on this to access more such competitive quizzes on Calculus.

Definite integrals Quiz 2

mathematics

 Determine the integral  $\dfrac{1}{4}\int\limits_{0}^{2\pi }{{{\cos }^{2}}\theta d\theta}$

 Solve the integral by simplifying the terms by applying ${{\cos }^{2}}\theta = \dfrac{1+\cos 2\theta }{2}$, then integrating within the given limit and put the maximum and minimum limit to evaluate the integral.

 $I=\dfrac{1}{4}\int\limits_{0}^{2\pi }{{{\cos }^{2}}\theta d\theta } \\\Rightarrow \dfrac{1}{4}\int\limits_{0}^{2\pi }{\left( \dfrac{1+\cos 2\theta }{2} \right)d\theta } \\\Rightarrow \dfrac{1}{8}{{\left[ \theta +\dfrac{\sin 2\theta }{2} \right]}_{0}}^{2\pi } \\\Rightarrow \dfrac{1}{8}\times 2\pi  \\ \Rightarrow \dfrac{\pi }{4} $ 

 Determine the value of $\int\limits_{1}^{2}{\dfrac{2{{x}^{5}}-x+3}{{{x}^{2}}}}$

 Solve the integral by separating the equation and then integrating within the given limits to get the value of integration.

  $I=\int\limits_{1}^{2}{\dfrac{2{{x}^{5}}-x+3}{{{x}^{2}}}}dx \\\Rightarrow \int\limits_{1}^{2}{2{{x}^{3}}}-\dfrac{1}{x}+\dfrac{3}{{{x}^{2}}}dx \\ \Rightarrow {{\left[ \dfrac{{{x}^{4}}}{2}-\ln \left| x \right|-\dfrac{3}{x} \right]}_{1}}^{2} \\\Rightarrow \dfrac{15}{2}-\ln \left| 2 \right|+\dfrac{3}{2} \\ \Rightarrow 9-\ln \left| 2 \right| $ 

 Find the constant after the integration of $\int\limits_{-1}^{1}{\dfrac{2{{x}^{2}}}{1+{{x}^{2}}}dx}$

 Solve the integral by adding $+1$ and $-1$ in the numerator and then simplifying the terms and integrating within the given limits to get the value.

  $I=\int\limits_{-1}^{1}{\dfrac{2{{x}^{2}}}{1+{{x}^{2}}}dx} \\ =2\int _{-1}^1\dfrac{x^2}{1+x^2}dx \\ =2\int _{-1}^1-\dfrac{1}{1+x^2}+1dx \\ =2\left(-\int _{-1}^1\dfrac{1}{1+x^2}dx+\int _{-1}^11dx\right) \\ =2\left(-\dfrac{\pi }{2}+2\right) \\ =-\pi +4$ 

 Find the value of numerator of $\int\limits_{0}^{1}{\sqrt[3]{1+7x}\text{ }dx}$

 Use the method of substitution to solve the integral, substitute $t=1+7x,dt=7dx$ and then solve the integral.

  $I=\int\limits_{0}^{1}{\sqrt[3]{1+7x}\text{ }dx} \\ \text{Let }t=1+7x,dt=7dx \\  \Rightarrow \dfrac{1}{7}\int\limits_{1}^{8}{{{t}^{1/3}}}dt \\ \Rightarrow \dfrac{1}{7}\times \dfrac{3}{4}\times {{\left[ {{t}^{4/3}} \right]}_{1}}^{8} \\ \Rightarrow \dfrac{3}{28}\times 15 \\ \Rightarrow \dfrac{45}{28} \\ \text{So, the numerator is }45$ 

 Find the difference between the constants of the integrals $\int\limits_{4}^{5}{\dfrac{{{x}^{3}}-3{{x}^{2}}-9}{{{x}^{3}}-3{{x}^{2}}}dx\ \text{  }}$ and $\int\limits_{0}^{1}{\dfrac{{{x}^{2}}+5x}{4x}}dx$

 First solve the individual integral by separating the terms at the operations. Then integrate the integrals separately and simplify the terms.Later evaluate the difference between the two integrals.

  ${{I}_{1}}=\int\limits_{4}^{5}{\dfrac{{{x}^{3}}-3{{x}^{2}}-9}{{{x}^{3}}-3{{x}^{2}}}dx\ \text{  }} \\\Rightarrow \int\limits_{4}^{5}{1-\dfrac{9}{{{x}^{3}}-3{{x}^{2}}}dx\ \text{  }} \\ =-\int _4^5\dfrac{9}{x^3-3x^2}dx+\int _4^51dx \\ = -\left(-\dfrac{3}{20}-\ln \left(5\right)+3\ln \left(2\right)\right)+1 \\ =\ln \left(5\right)-3\ln \left(2\right)+\dfrac{23}{20} \\ {{I}_{2}}=\int\limits_{0}^{1}{\dfrac{{{x}^{2}}+5x}{4x}}dx \\ =\dfrac{1}{4}\cdot \int _0^1\dfrac{x^2+5x}{x}dx \\ =\dfrac{1}{4}\left(\int _0^1\dfrac{x^2}{x}dx+\int _0^1\dfrac{5x}{x}dx\right) \\ =\dfrac{1}{4}\left(\dfrac{1}{2}+5\right) \\ =\dfrac{11}{8} \\ \text{Difference = }\ln \left(5\right)-3\ln \left(2\right)+\dfrac{23}{20}-\dfrac{11}{8} = \ln \left(\dfrac{5}{8}\right)-\dfrac{9}{40} $ 