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    Integration as an Inverse Process of Differentiation - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • Integration as Reverse Process of Differentiation is considered one the most difficult concept.

    • 68 Questions around this concept.

    Solve by difficulty

    $\int \frac{(1+x)^2}{\sqrt{x}} d x$ equals

    $\int \frac{\sec ^2 \mathrm{x}}{1+\tan \mathrm{x}} \mathrm{dx}$ equals

    $\int \frac{1}{(x+1) \sqrt{x^2-1}} d x$ equals

    $\int \frac{2^x+3^x}{4^x} d x$ equals

    $\int \frac{x^{2}+ \cos^2x}{(1+x^{2}) \sin^{2}x}.dx$

    is equal to?

    $\int \frac{x-\sin x}{1-\cos x}dx =$

    The value of $\int_0^{2 \pi} \min \left(|x-\pi|, \cos ^{-1}(\cos x)\right) d x$ is

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    Concepts Covered - 1

    Integration as Reverse Process of Differentiation

    Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. 

    For example,

    $\begin{aligned} & \frac{d}{d x}(\sin x)=\cos x \\ & \frac{d}{d x}\left(x^2\right)=2 x \\ & \frac{d}{d x}\left(e^x\right)=e^x\end{aligned}$

    In the above example, the function cos(x) is the derivative of sin(x). We say that sin(x) is an anti-derivative (or an integral) of cos(x). Similarly, x2 and eare the antiderivatives (or integrals) of 2x and ex respectively.  

    Also note that the derivative of a constant  (C) is zero. So we can write the above examples as:

    $\begin{aligned} & \frac{d}{d x}(\sin x+c)=\cos x \\ & \frac{d}{d x}\left(x^2+c\right)=2 x \\ & \frac{d}{d x}\left(e^x+c\right)=e^x\end{aligned}$

    Thus, the anti-derivatives (or integrals) of the above functions are not unique. Actually, there exist infinitely many anti-derivatives of each of these functions which can be obtained by selecting C arbitrarily from the set of real numbers. 

    For this reason, C is referred to as an arbitrary constant. In fact, C is the parameter by varying which one gets different anti-derivatives (or integrals) of the given function. 

    If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx. 

    By definition,

    $\int \mathrm{f}(\mathrm{x}) \mathrm{dx}=\mathrm{F}(\mathrm{x})+\mathrm{c}, \quad$ where $\mathrm{F}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})$ and ' $\mathrm{c}^{\prime}$ is constant.

    Here, 

    \begin{array}{c||c} \hline\\ \mathbf { Symbols/Terms/Phrases } & \mathbf{Meaning} \\\\ \hline \hline \\\int f(x) d x & {\text { Integral of } f \text { with respect to } x} \\\\ \hline \\f(x) \text { in } \int f(x) d x & {\text { Integrand }} \\\\\hline \\ x \text { in } \int f(x) d x & {\text { Variable of integration }} \\\\\hline \\ \text{An integral of f } & {\text { A function F such that F'(x) = f(x)}}\\\\\hline \end{array}

    Rules of integration

    f(x) and g(x) are functions with antiderivatives ∫ f(x) and  ∫ g(x) dx. Then,

    (a) $\int \mathrm{kf}(x) d x=k \int f(x) d x$ for any constant $k$.
    (b) $\int(f(x)+g(x)) d x=\int f(x) d x+\int g(x) d x$
    (c) $\int(f(x)-g(x)) d x=\int f(x) d x-\int g(x) d x$

    Standard Integration Formulae

    Since,$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{F}(\mathrm{x}))=\mathrm{f}(\mathrm{x}) \Leftrightarrow \int \mathrm{f}(\mathrm{x}) \mathrm{dx}=\mathrm{F}(\mathrm{x})+\mathrm{c}$

    Based on this definition and various standard formulas (which we studied in Limit, Continuity, and Differentiability) we can obtain the following important integration formulae, 

    1.$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{kx})=\mathrm{k}\Rightarrow\int\mathrm{kdx}=\mathrm{kx}+\mathrm{C}$, where k is a constant
    2.$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}\right)=\mathrm{x}^{\mathrm{n}}, \mathrm{n} \neq-1 \Rightarrow \int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}, \mathrm{n} \neq-1$
    3. $\frac{\mathrm{d}}{\mathrm{dx}}(\log |\mathrm{x}|)=\frac{1}{\mathrm{x}} \Rightarrow \int \frac{1}{\mathrm{x}} \mathrm{dx}=\log |\mathrm{x}|+\mathrm{C}$, when $\mathrm{x} \neq 0$
    4. $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{e}^{\mathrm{x}} \Rightarrow \int \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^{\mathrm{x}}+\mathrm{C}$
    5.$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{a}^{\mathrm{x}}}{\log_{\mathrm{e}}\mathrm{a}}\right)=\mathrm{a}^{\mathrm{x}}, \mathrm{a}>0, \mathrm{a} \neq 1 \Rightarrow \int \mathrm{a}^{\mathrm{x}} \mathrm{dx}=\frac{\mathrm{a}^{\mathrm{x}}}{\log _{\mathrm{e}} \mathrm{a}}+\mathrm{C}$

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    Reference Books

    Integration as Reverse Process of Differentiation

    Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

    Page No. : 7.1

    Line : 1

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