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# integration of root x

The integral of the root x, can be found using power rule. The integral of $\sqrt x$ with respect to (x) is:

$\int \sqrt x \, dx =\frac{2}{3} x^{3/2} + C$

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant.

## Proof of integration of root x

To integrate the square root of $x$ (i.e., $\sqrt{x}$), we express it as $x^{1/2}$. The integral is then:

$$\int \sqrt{x} \, dx = \int x^{1/2} \, dx$$

To integrate this, we use the power rule for integration:

$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$

where $C$ is the constant of integration. Applying this rule:

$$\int x^{1/2} \, dx = \frac{x^{(1/2) + 1}}{(1/2) + 1} + C = \frac{x^{3/2}}{3/2} + C$$

Simplifying the fraction:

$$= \frac{2}{3} x^{3/2} + C$$

So, the integral of (\sqrt{x}$is: $$\frac{2}{3} x^{3/2} + C$$ ## Integration based on root x $$\int \sqrt{ax + b} \, dx = \frac{2}{3a} (ax + b)^{3/2} + C$$ Proof Choose a substitution that simplifies the integrand. Let ($u = ax + b$). Then, ($du = a \, dx$) or ($dx = \frac{du}{a}$). Substitute ( u ) and ( dx ) into the integral: $$= \int \sqrt{u} \cdot \frac{du}{a}$$ $$= \frac{1}{a} \int u^{1/2} \, du$$ $$= \frac{1}{a} \cdot \frac{2}{3} u^{3/2} + C$$ Here, ( C ) is the constant of integration. Substitute back for ( u ) to get the final answer: $$\int \sqrt{ax + b} \, dx= \frac{2}{3a} (ax + b)^{3/2} + C$$ ## Definite Integral of root x To evaluate a definite integral of the root x,$\int_a^b \sqrt x \, dx$, where$ a $and$ b $are the limits of integration, we follow a similar process as with the indefinite integral and the we’ll apply the limits at the end. $$\int_a^b \sqrt(x) \, dx = [ \frac{2}{3} x^{3/2}]_a^b$$ ## Solved Examples Question 1 $$\int \sqrt{3x – 1} \, dx$$ Solution Choose a substitution that simplifies the integrand. Let$ u = 3x – 1 $. Then,$ du = 3 \, dx $or$ dx = \frac{du}{3} $. Substitute$ u $and$ dx $into the integral: $$=\int \sqrt{u} \cdot \frac{du}{3}$$ You can simplify this to: $$=\frac{1}{3} \int u^{1/2} \, du$$ Now integrate with respect to$ u $: $$=\frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C$$ Here,$ C $is the constant of integration. Substitute back for$ u $to get the final answer: $$= \frac{2}{9} (3x – 1)^{3/2} + C$$ Question 2 $$\int \sqrt{x} \cdot \ln(x) \, dx$$ Solution Integrating$ \sqrt{x} \cdot \ln(x) $requires the use of integration by parts. $$\int u \, dv = uv – \int v \, du$$ • Let$ u = \ln(x) $(since the derivative of$ \ln(x) $is simpler than its integral). • Then$ dv = \sqrt{x} \, dx $(since we know how to integrate$ \sqrt{x} $). Then •$ du = \frac{1}{x} \, dx $. •$ v = \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} \$ (using the power rule).

Therefore

$$\int \sqrt{x} \cdot \ln(x) \, dx = uv – \int v \, du$$
$$= \ln(x) \cdot \frac{2}{3} x^{3/2} – \int \frac{2}{3} x^{3/2} \cdot \frac{1}{x} \, dx$$
$$= \frac{2}{3} x^{3/2} \ln(x) – \frac{2}{3} \int x^{1/2} \, dx$$
$$= \frac{2}{3} x^{3/2} \ln(x) – \frac{2}{3} \cdot \frac{2}{3} x^{3/2} + C$$
$$= \frac{2}{3} x^{3/2} \ln(x) – \frac{4}{9} x^{3/2} + C$$

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