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
$$ \int \sqrt{x} \cdot \ln(x) \, dx = uv – \int v \, du $$
$$ = \ln(x) \cdot \frac{2}{3} x^{3/2} – \int \frac{2}{3} x^{3/2} \cdot \frac{1}{x} \, dx $$
$$ = \frac{2}{3} x^{3/2} \ln(x) – \frac{2}{3} \int x^{1/2} \, dx $$
$$ = \frac{2}{3} x^{3/2} \ln(x) – \frac{2}{3} \cdot \frac{2}{3} x^{3/2} + C $$
$$ = \frac{2}{3} x^{3/2} \ln(x) – \frac{4}{9} x^{3/2} + C $$
I hope you like this article on integration of root x interesting and useful
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