If f(x)=1 then

$\int_{}^{} k.dx = k\int_{}^{}1.dx = k\int_{}^{}x^{0}dx = kx +c$

Thus, integration of a constant k with respect to x is kx.

The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.

$\int_{}^{} (f_{1}(x) \pm f_{2}(x)....\pm f_{n}(x))dx = \int_{}^{}f_{1}(x) dx \pm \int_{}^{}f_{2}(x)dx \pm .... \pm \int_{}^{}f_{n}(x)dx$

Using the property of constant of integration

$\int_{}^{} k.dx = k \int_{}^{} k.dx $ = kx + c, where c is the constant of integration

In the given question k= 2

$\int_{}^{} 2 dx = 2\int_{}^{} dx $ = 2x + c

**Example 2 : ** Evaluate the integration of $5x^{2}$ with respect to x.

**Solution :** $\int_{}^{} 5x^{2} dx$

Here we will use the property of constant of integration and power rule of integration

$\int_{}^{} k.dx = k \int_{}^{} k.dx $ = kx + c, where c is the constant of integration

$\int_{}^{} x^{n} dx= \frac{x^{n+1}}{n+1} + c$

In the given question k= 5

$\int_{}^{} 5x^{2} dx = 5\int_{}^{}x^{2}dx $

= 5$\int_{}^{}x^{2}dx $

= 5 $\frac{x^{2+1}}{2+1}$

$\int_{}^{} 5x^{2} dx = \frac {5x^{3}}{3} + c $

**Example 3:** Evaluate the integration of $x^{3} - 3x^{2} + 5x + 2$ with respect to x.

**Solution :** $\int_{}^{}(x^{3} - 3x^{2} + 5x + 2) dx $

Here we will use the power rule, constant rule and 2nd property of integration

$\int_{}^{}(x^{3} - 3x^{2} + 5x + 2) dx = \int_{}^{}x^{3}dx -\int_{}^{}3x^{2}dx + \int_{}^{}5x dx + \int_{}^{}2 dx $

= $\frac{x^{3+1}}{3+1} - \frac{3x^{2+1}}{2+1} + \frac{5x^{1+1}}{1+1} + 2x $

= $\frac{x^{4}}{4} - \frac{3x^{3}}{3} + \frac{5x^{2}}{2} + 2x $

$\int_{}^{}(x^{3} - 3x^{2} + 5x + 2) dx = \frac{x^{4}}{4} - x^{3} + \frac{5x^{2}}{2} + 2x$ +c

**Example 4:** Evaluate the integration of $x^{3} - \sqrt{x}$ with respect to x.

**Solution :** $\int_{}^{}(x^{3} - \sqrt{x}) dx $

Here we will use the power rule and 2nd property of integration

$\int_{}^{}(x^{3} - \sqrt{x}) dx = \int_{}^{}x^{3}dx -\int_{}^{}\sqrt{x} dx $

= $\int_{}^{}x^{3}dx -\int_{}^{}x^{1/2} dx$

= $\frac{x^{3+1}}{3+1} - \frac{x^(\frac{1}{2}+1)}{\frac{1}{2}+1} $

= $\frac{x^{4}}{4} - \frac{x^(\frac{3}{2})}{\frac{3}{2}} $

$\int_{}^{}(x^{3} - \sqrt{x}) dx =\frac{x^{4}}{4} - \frac{2\sqrt{x^{3}}}{3}$ + c

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