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Integration as Reverse Process of Differentiation is considered one the most difficult concept.
44 Questions around this concept.
The integral value of is
The integral of is
where C is a constant of integration, then the value of k is equal to
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If dx and I (0) = 1, then is equal to :
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 ex are 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,
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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