Evaluate the integral: I = integral sqrt(2x - 3) dx
Evaluate the integral: $$ I = \int \sqrt{2x - 3} \, dx. $$
Explanation of Method
Square roots of linear expressions are handled the same way as other powers of linear expressions: substitute $u$ for the inside of the root, rewrite the root as a fractional power $u^{1/2}$, and apply the power rule before substituting back.
- Substitute $u$ for the linear expression inside the root.
- Rewrite the square root as a fractional power of $u$.
- Integrate using the power rule.
- Substitute back and add the constant of integration.
Step 1 · Use substitution
To simplify the integral, we will use substitution. Let:
$$ u = 2x - 3. $$
Then, differentiate $u$ with respect to $x$:
$$ \frac{du}{dx} = 2 \quad \Rightarrow \quad dx = \frac{du}{2}. $$
Substituting into the integral, we get:
$$ I = \int \sqrt{u}\cdot \frac{du}{2} = \frac{1}{2}\int \sqrt{u}\,du. $$
Step 2 · Integrate
We can now integrate $\sqrt{u} = u^{1/2}$. Using the power rule for integration:
$$ \int u^n\,du = \frac{u^{n+1}}{n+1} + C \quad \text{for } n\neq -1, $$
we get:
$$ \int u^{1/2}\,du = \frac{u^{3/2}}{3/2} = \frac{2}{3}u^{3/2}. $$
Step 3 · Substitute back $u = 2x - 3$
Now, substitute back $u = 2x-3$ into the result:
$$ I = \frac{1}{2}\cdot \frac{2}{3}(2x-3)^{3/2}. $$
Simplifying the constants:
$$ I = \frac{1}{3}(2x-3)^{3/2}. $$
Step 4 · Add the constant of integration
Finally, we add the constant of integration $C$ to the result:
$$ I = \frac{1}{3}(2x-3)^{3/2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 2x-3$ |
| $dx$ in terms of $du$ | $\tfrac{1}{2}du$ |
| Antiderivative in $u$ | $\tfrac{2}{3}u^{3/2}$ |
Additional Notes
- The domain of $I$ requires $x \geq \tfrac{3}{2}$, since $2x-3 \geq 0$ is needed for the square root to be real.
- The two constant factors — $\tfrac12$ from $dx=\tfrac{du}{2}$ and $\tfrac23$ from the power rule — multiply to give the final coefficient $\tfrac13$.
- The result can be checked by differentiating $\tfrac13(2x-3)^{3/2}$ with the chain rule and confirming it returns $\sqrt{2x-3}$.
Result
The solution to the integral is: