The integral in the image is:

$\int_{0}^{3} \sqrt{x^3 + 3x^2 + 3x + 1} \, dx$

Let's analyze and simplify this expression.

Step 1: Recognize the Expression Inside the Square Root

The expression under the square root, $x^3 + 3x^2 + 3x + 1$, is a cubic polynomial. It resembles the binomial expansion of $(x + 1)^3$, which expands as follows: $(x + 1)^3 = x^3 + 3x^2 + 3x + 1$

Therefore, we can rewrite the integrand as: $\sqrt{x^3 + 3x^2 + 3x + 1} = \sqrt{(x + 1)^3} = |x + 1|^{3/2}$

Since $x$ ranges from $0$ to $3$, $x + 1$ will always be positive over this interval, so we can drop the absolute value: $\int_{0}^{3} (x + 1)^{3/2} \, dx$

Step 2: Integrate $(x + 1)^{3/2}$

Now we need to find: $\int (x + 1)^{3/2} \, dx$

Using the power rule for integration, where $\int u^n \, du = \frac{u^{n+1}}{n+1}$, let $u = x + 1$. Then $du = dx$.

The integral becomes: $\int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}$

Substituting back $u = x + 1$, we get: $\int (x + 1)^{3/2} \, dx = \frac{2}{5} (x + 1)^{5/2}$

Step 3: Evaluate the Definite Integral

Now we evaluate from $x = 0$ to $x = 3$: $\int_{0}^{3} (x + 1)^{3/2} \, dx = \frac{2}{5} \left[ (x + 1)^{5/2} \right]_{0}^{3}$

Substitute the limits: $= \frac{2}{5} \left[ (3 + 1)^{5/2} - (0 + 1)^{5/2} \right]$ $= \frac{2}{5} \left[ 4^{5/2} - 1^{5/2} \right]$

Calculate $4^{5/2}$: $4^{5/2} = (2^2)^{5/2} = 2^5 = 32$

And $1^{5/2} = 1$.

So: $= \frac{2}{5} (32 - 1) = \frac{2}{5} \cdot 31 = \frac{62}{5} = 12.4$

Final Answer

$\int_{0}^{3} \sqrt{x^3 + 3x^2 + 3x + 1} \, dx = 12.4$

Would you like more details on any of these steps?

Related Questions

1. How does the binomial expansion simplify the process of integrating polynomials?
2. Can we apply substitution in more complex integrals with similar polynomials?
3. What are some techniques to handle integrals with higher powers inside a square root?
4. How does recognizing a perfect power simplify square root integrals?
5. How would this integral change if the bounds or exponents were different?

Tip: Recognizing binomial expansions in integrals can save time by simplifying complex polynomials into simpler terms.