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Integrals

Integration is an inverse process of differentiation. In this chapter, we will learn how to find the integral of a function. Its knowledge in calculus is very much needed for finding the areas under curves, etc. This chapter consists of integration of a variety of functions by substitutions, by parts and by partial fractions. Definite integrals as a limit of a sum , fundamental theorem of calculus, properties of definite integrals.

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Exercise 1

  • Q1 Integrals sin 2x
    Ans:

    The anti derivative of sin 2x is a function of x whose derivative is sin 2x.

    It is known that,

    \begin{align} \frac {d}{dx} (cos 2x) = 2 sin2x \end{align}

    ⇒ \begin{align} sin 2x =-\frac {1}{2} \frac {d}{dx}(cos 2x) \end{align} 

    ∴  \begin{align} sin 2x = \frac {d}{dx}\left(-\frac {1}{2}cos 2x\right) \end{align} 

    Therefore, the anti derivative of sin2x is 

    \begin{align} sin 2x \;is -\frac {1}{2}cos 2x \end{align} 


    Q2 Integrals cos3x
    Ans:

    The anti derivative of cos 3x is a function of x whose derivative is cos 3x.

    It is known that,

    \begin{align} \frac {d}{dx} (sin 3x) = 3 cos3x \end{align}

    ⇒ \begin{align} cos 3x =\frac {1}{3} \frac {d}{dx}(sin 3x) \end{align} 

    ∴  \begin{align} cos 3x = \frac {d}{dx}\left(\frac {1}{3}sin 3x\right) \end{align} 

    Therefore, the anti derivative of cos3x is 

    \begin{align} sin 2x \;is \frac {1}{3}sin 3x \end{align} 


    Q3 Integrals e2x
    Ans:

    The anti derivative of e2x is a function of x whose derivative is e2x.

    It is known that,

    \begin{align} \frac {d}{dx} (e^{2x}) = 2e^{2x} \end{align}

    ⇒ \begin{align} e^{2x} =\frac {1}{2} \frac {d}{dx}(e^{2x}) \end{align} 

    ∴  \begin{align} e^{2x} = \frac {d}{dx}\left(\frac {1}{2}e^{2x}\right) \end{align} 

    Therefore, the anti derivative of e2x

    \begin{align} e^{2x} \;is \frac {1}{2}e^{2x} \end{align} 

     


    Q4 Integrals (ax + b)2
    Ans:

    The anti derivative of (ax + b)2 is a function of x whose derivative is (ax + b)2.

    It is known that,

    \begin{align} \frac {d}{dx} ((ax+b)^3) = 3a(ax+b)^2 \end{align}

    ⇒ \begin{align} (ax + b)^2 =\frac {1}{3a} \frac {d}{dx}(ax+b)^3 \end{align} 

    ∴  \begin{align} (ax + b)^2 = \frac {d}{dx}\left(\frac {1}{3a}(ax + b)^3\right) \end{align} 

    Therefore, the anti derivative of (ax +b)2

    \begin{align} (ax + B)^2 \;is \frac {1}{3a}(ax +b)^3 \end{align}


    Q5 sin 2x – 4e3x
    Ans:

    The anti derivative of sin 2x – 4e3x is the function of x whose derivative is sin 2x – 4e3x.

    It is known that,

    \begin{align} \frac {d}{dx} \left(-\frac{1}{2}cos 2x  –  \frac {4}{3} e^{3x}\right) = sin2x  –  4e^{3x} \end{align} 

    Therefore, the anti derivative of (sin 2x – 4e3x) is \begin{align}  \left(-\frac{1}{2}cos 2x  –  \frac {4}{3} e^{3x}\right) \end{align} 

     

     


    Q6 \begin{align} \int \left(4e^{3x} + 1\right).dx \end{align}
    Ans:

    \begin{align} \int \left(4e^{3x} + 1\right).dx \end{align}

    \begin{align} =4\int e^{3x}.dx + \int 1.dx \end{align}

    \begin{align} =4\left(\frac {e^{3x}}{3}\right) + x + C \end{align}

    \begin{align} =\frac {4}{3} e^{3x} + x + C \end{align}


    Q7 \begin{align} \int x^2\left(1 - \frac{1}{x^2}\right).dx \end{align}
    Ans:

    \begin{align} \int x^2\left(1 - \frac{1}{x^2}\right).dx \end{align}

     \begin{align} =\int \left(x^2 - 1\right).dx \end{align}

     \begin{align} =\int x^2.dx - \int1.dx \end{align}

     \begin{align} =\frac{x^3}{3} - x + C \end{align}


    Q8 \begin{align} \int \left({a}{x^2} + bx + c\right) .dx\end{align}
    Ans:

    \begin{align} \int \left({a}{x^2} + bx + c\right) .dx\end{align}

    \begin{align} =a\int {x^2}.dx + b\int x.dx + c\int 1.dx \end{align}

    \begin{align} =a\left(\frac {x^3}{3}\right) + b\left(\frac {x^2}{2}\right) + cx + C\end{align}

     


    Q9 \begin{align} \int \left({2}{x^2} + e^x\right) .dx\end{align}
    Ans:

    \begin{align} \int \left({2}{x^2} + e^x\right) .dx\end{align}

    \begin{align} =2\int {x^2}.dx + \int e^x.dx \end{align}

    \begin{align} =2\left(\frac {x^3}{3}\right) +e^x + C\end{align}

    \begin{align} =\frac {2}{3}.x^3 +e^x + C\end{align}


    Q10 \begin{align} \int \left(\sqrt{x} - \frac {1}{\sqrt{x}}\right)^2 .dx\end{align}
    Ans:

    \begin{align} \int \left(\sqrt{x} - \frac {1}{\sqrt{x}}\right)^2 .dx\end{align}

    \begin{align} =\int \left(x + \frac {1}{x} - 2\right) .dx\end{align}

    \begin{align} =\int x .dx + \int \frac {1}{x} . dx - 2\int 1. dx\end{align}

    \begin{align} =\frac{x^2}{2} + \log\left(\left|x\right|\right) - 2x + C\end{align}

     


    Q11 \begin{align} \int \frac{x^3 + 5x^2 - 4}{x^2} . dx\end{align}
    Ans:

    \begin{align} \int \frac{x^3 + 5x^2 - 4}{x^2} . dx\end{align}

    \begin{align} =\int \left(x + 5 - 4x^{-2}\right) . dx\end{align}

    \begin{align} =\int x .dx + 5 \int 1.dx- 4 \int x^{-2} .dx\end{align}

    \begin{align} =\frac {x^2}{2} + 5x + 4 \left(\frac{x^{-1}}{-1}\right) + C\end{align}

    \begin{align} =\frac {x^2}{2} + 5x + \frac{4}{x} + C\end{align}


    Q12 \begin{align} \int \frac{x^3 + 3x + 4}{\sqrt{x}} . dx\end{align}
    Ans:

    \begin{align} \int \frac{x^3 + 3x + 4}{\sqrt{x}} . dx\end{align}

    \begin{align} =\int \left(x^\frac{5}{2} + 3x^\frac{1}{2} + 4x^\frac{1}{2}\right) . dx\end{align}

    \begin{align} =\frac{\left(x^{\displaystyle\frac72}\right)}{\displaystyle\frac72}+ \frac{3\left(x^{\displaystyle\frac32}\right)}{\displaystyle\frac32} + \frac{4\left(x^{\displaystyle\frac12}\right)}{\displaystyle\frac12} + C\end{align}

    \begin{align} =\frac27\left(x^\frac72\right)+ 2\left(x^\frac32\right) + 8\left(x^\frac12\right) + C\end{align}

    \begin{align} =\frac27\left(x^\frac72\right)+ 2\left(x^\frac32\right) + 8\left(\sqrt x\right) + C\end{align}

     


    Q13 \begin{align} \int \frac{x^3 - x^2 + x - 1}{x-1} . dx\end{align}
    Ans:

    \begin{align} \int \frac{x^3 - x^2 + x - 1}{x-1} . dx\end{align}

    On dividing, we obtain

    \begin{align} =\int \left({x^2 + 1}\right) . dx \end{align}

    \begin{align} =\int {x^2} . dx + \int 1 .dx \end{align}

    \begin{align} =\frac {x^3}{3}  + x + C \end{align}


    Q14 \begin{align} \int\left(1-x\right).\sqrt {x}.dx\end{align}
    Ans:

    \begin{align} \int\left(1-x\right).\sqrt {x}.dx\end{align}

    \begin{align} =\int \left(\sqrt {x} - x^\frac32\right).dx\end{align}

    \begin{align} =\frac{x^{\displaystyle\frac32}}{\displaystyle\frac32} - \frac{x^{\displaystyle\frac52}}{\displaystyle\frac52}+C \end{align}

    \begin{align} =\frac23\;x^\frac32\; - \frac25\;x^\frac52\;+C \end{align}

     

     


    Q15 \begin{align} \int\sqrt {x}.\left(3x^2+2x + 3\right).dx\end{align}
    Ans:
    \begin{align} \int\sqrt {x}.\left(3x^2+2x + 3\right).dx\end{align}
     
    \begin{align} =\int\left(3x^\frac52+2x^\frac32 + 3x^\frac12\right).dx\end{align}
     
    \begin{align} =3\int x^\frac52 .dx+2 \int x^\frac32 .dx + 3\int x^\frac12.dx\end{align}
     
    \begin{align} =3\left(\frac{x^{\displaystyle\frac72}}{\displaystyle\frac72}\right) + 2\left(\frac{x^{\displaystyle\frac52}}{\displaystyle\frac52}\right)+3\left(\frac{x^{\displaystyle\frac32}}{\displaystyle\frac32}\right)+C \end{align}
     
     \begin{align} =\frac67\;x^\frac72\; + \frac45\;x^\frac52\;+ 2x^\frac32\; +C \end{align}
     

    Q16 \begin{align} \int\left(2x - 3Cosx + e^x\right).dx\end{align}
    Ans:

    \begin{align} \int\left(2x - 3Cosx + e^x\right).dx\end{align}

    \begin{align} =2\int x .dx - 3\int Cosx .dx + \int e^x.dx\end{align}

    \begin{align} =\frac{2x^2}{2} - 3(Sinx) +  e^x + C\end{align}

    \begin{align} =x^2 - 3Sinx +  e^x + C\end{align}

     


    Q17 \begin{align} \int\left(2x^2-3Sinx +5\sqrt {x}\right).dx\end{align}
    Ans:

    \begin{align} \int\left(2x^2-3Sinx +5\sqrt {x}\right).dx\end{align}

    \begin{align} =2\int x^2.dx-3\int Sinx.dx +5\int x^\frac12.dx \end{align}

    \begin{align} =\frac{2x^3}{3} - 3(- Cos x) +5\left(\frac{x^{\displaystyle\frac32}}{\displaystyle\frac32}\right) + C \end{align}

    \begin{align} =\frac{2}{3}.x^3 + 3Cos x + \frac{10}{3}.x^\frac{3}{2} + C\end{align}

     


    Q18 \begin{align} \int sec x . \left(sec x + tan x\right) .dx \end{align}
    Ans:

    \begin{align} \int sec x . \left(sec x + tan x\right) .dx \end{align}

    \begin{align} =\int \left(sec^2 x  + sec x . tan x\right) .dx \end{align}

    \begin{align} =\int sec^2 x . dx + \int sec x . tan x .dx \end{align}

    \begin{align} = tan x  + sec x + C \end{align}


    Q19 \begin{align} \int \frac {sec^2 x}{Coses^2 x} . dx\end{align}
    Ans:

    \begin{align} \int \frac {sec^2 x}{Coses^2 x} . dx\end{align}

    \begin{align} =\int \left(\frac{\frac {1}{Cos^2 x}}{\frac{1}{sin^2 x}}\right) . dx\end{align}

    \begin{align} =\int \left(\frac{Sin^2x}{Cos^2x}\right) . dx\end{align}

    \begin{align} =\int tan^2 x . dx\end{align}

    \begin{align} =\int \left(sec^2x - 1\right) . dx\end{align}

    \begin{align} =\int sec^2x . dx - \int 1. dx\end{align}

    \begin{align} = tanx - x + C\end{align}


    Q20 \begin{align} \int \left(\frac {2-3sin x}{cos^2 x}\right) . dx\end{align}
    Ans:

    \begin{align} \int \left(\frac {2-3sin x}{cos^2 x}\right) . dx\end{align}

    \begin{align} =\int \left(\frac {2}{cos^2 x} - \frac{3sinx}{cos^2x}\right) . dx\end{align}

    \begin{align} =2\int sec^2 x .dx   - 3 \int tan x . sec x. dx\end{align}

    \begin{align} =2 tan x - 3 sec x + C\end{align} 


    Q21 The anti derivative of \begin{align} \left(\sqrt x + \frac {1}{\sqrt x}\right)\end{align} equals to \begin{align} (A) \frac{1}{3}.x^\frac{1}{3} + 2.x^\frac{1}{2} +C \;\;\;\; (B) \frac{2}{3}.x^\frac{2}{3} + \frac{1}{2}.x^2 +C \end{align}
    \begin{align} (C) \frac{2}{3}.x^\frac{3}{2} +2 x^\frac{1}{2} +C \;\;\;\;(D) \frac{3}{2}.x^\frac{3}{2} +\frac{1}{2}. x^\frac{1}{2} +C \end{align}
    Ans:

    \begin{align} \int\left(\sqrt x + \frac {1}{\sqrt x}\right).dx \end{align}

    \begin{align}= \int  x^\frac{1}{2}.dx + \int x^\frac{-1}{2}.dx \end{align}

    \begin{align}= \left(\frac {x^\frac{3}{2}}{\frac{3}{2}}\right) + \left(\frac {x^\frac{1}{2}}{\frac{1}{2}}\right) + C\end{align}

    \begin{align}= \frac {2}{3} . x^{\frac{3}{2}}+ 2x^\frac{1}{2} + C\end{align}

    Hence, the Correct Answer is C.


    Q22 If \begin{align} \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\end{align} such that f(2) = 0 , then f(x) is
    \begin{align} (A) x^4 + \frac {1}{x^3} - \frac{129}{8} \;\;\;\;(B) x^3 + \frac{1}{x^4} + \frac{129}{8}\end{align}
    \begin{align} (c) x^3 + \frac {1}{x^4} + \frac{129}{8} \;\;\;\;(D) x^3 + \frac{1}{x^4} - \frac{129}{8}\end{align}
    Ans:

    It is given that,

    \begin{align} \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\end{align}  

    ∴ Anti derivative of 

    \begin{align} 4x^3 - \frac{3}{x^4} = f(x)\end{align}  

    ∴ \begin{align} f(x)= \int \left(4x^3 - \frac{3}{x^4}\right).dx\end{align}  

    \begin{align} f(x)=  4\int x^3.dx - 3\int {x^{-4}}.dx\end{align}  

    \begin{align} f(x)= 4\left(\frac {x^4}{4}\right) - 3\left(\frac {x^{-3}}{-3}\right) + C\end{align}  

    ∴ \begin{align} f(x)=   x^4 + \frac{1}{x^3} + C\end{align}  

    Also, f(2) = 0

    ∴ \begin{align} f(2) =\left(2\right)^4 + \frac{1}{\left(2\right)^3} + C = 0 \end{align}

    => \begin{align} 16 + \frac{1}{8} + C = 0 \end{align}

    => \begin{align} C = -\left(16 + \frac{1}{8}\right) \end{align}

    => \begin{align} C = \frac{-129}{8} \end{align}

    ∴  \begin{align} f(x)= x^4 + \frac{1}{x^3} -\frac{129}{8}  \end{align}

    Hence, the correct answer is A.

     

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