To solve the given definite integral:

$\int_{0}^{3} \sqrt{1 + 7x} \, dx$

Step-by-Step Solution:

1. Substitution: Let $u = 1 + 7x$. Then: $du = 7 \, dx \quad \text{or} \quad dx = \frac{du}{7}.$

   Also, when $x = 0$, $u = 1$, and when $x = 3$, $u = 22$.

2. Rewrite the integral: Substitute $u$ and $dx$ into the integral: $\int_{0}^{3} \sqrt{1 + 7x} \, dx = \int_{1}^{22} \sqrt{u} \cdot \frac{1}{7} \, du.$

   This simplifies to: $\frac{1}{7} \int_{1}^{22} u^{1/2} \, du.$

3. Integrate: The integral of $u^{1/2}$ is: $\int u^{1/2} \, du = \frac{2}{3} u^{3/2}.$

   Therefore: $\frac{1}{7} \int_{1}^{22} u^{1/2} \, du = \frac{1}{7} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{22}.$

4. Evaluate the definite integral: First, compute the antiderivative at the bounds: $\frac{2}{3} u^{3/2} \Big|_{1}^{22} = \frac{2}{3} \left( 22^{3/2} - 1^{3/2} \right).$

   $22^{3/2} = \sqrt{22^3} = \sqrt{10648} \approx 103.29$, and $1^{3/2} = 1$.

   So: $\frac{2}{3} \left( 22^{3/2} - 1 \right) = \frac{2}{3} (103.29 - 1) = \frac{2}{3} (102.29).$

   Simplify further: $\frac{1}{7} \cdot \frac{2}{3} \cdot 102.29 = \frac{204.58}{21} \approx 9.74.$

Final Answer:

$\int_{0}^{3} \sqrt{1 + 7x} \, dx \approx 9.74$

Would you like a detailed verification or have additional questions?

1. What is the significance of substitution in integral calculations?
2. How does changing limits of integration simplify solving definite integrals?
3. Can this problem be solved numerically for higher precision?
4. What are the applications of such integrals in physics or engineering?
5. How would the solution change if the integral were indefinite?

Tip: Always double-check your limits of integration when performing substitutions.