To find the definite integral of the function $y = x^2 + 4x - 3$ at $x = 5$, we calculate the integral of the function from 0 to 5. The definite integral of a function $f(x)$ from $a$ to $b$ is:

$\int_a^b f(x) \, dx$

Here, $f(x) = x^2 + 4x - 3$, and we want to integrate from 0 to 5.

Step 1: Find the indefinite integral

The indefinite integral of $f(x) = x^2 + 4x - 3$ is:

$\int (x^2 + 4x - 3) \, dx = \frac{x^3}{3} + 2x^2 - 3x + C$

Step 2: Apply the limits

Now, we evaluate this at the limits 0 and 5:

$\left[ \frac{x^3}{3} + 2x^2 - 3x \right]_0^5$

At $x = 5$:

$\frac{5^3}{3} + 2(5^2) - 3(5) = \frac{125}{3} + 2(25) - 15 = 41.67 + 50 - 15 = 76.67$

At $x = 0$:

$\frac{0^3}{3} + 2(0^2) - 3(0) = 0$

Step 3: Calculate the definite integral

Now, subtract the result at $x = 0$ from the result at $x = 5$:

$76.67 - 0 = 76.67$

Final Answer

The value of the definite integral from 0 to 5 of $x^2 + 4x - 3$ is approximately $76.7$ to the nearest 0.1.

Would you like more details on any step or have any questions?

Here are five related questions to explore:

1. What is the definite integral from 1 to 5 of $x^2 + 4x - 3$?
2. How do you find the area under a curve between two points?
3. How does changing the limits of integration affect the result?
4. What is the difference between definite and indefinite integrals?
5. How would the integral change if the function was $x^2 + 4x + 3$ instead?

Tip: The definite integral represents the area under the curve between two points on the x-axis.