Let's solve each of the problems listed in the image by computing the area under the curves using definite integrals.

Problem 1: $y = x^2 + x - 12$ and the x-axis

We are asked to find the area between the curve and the x-axis. The area can be found using a definite integral. First, we find the points where the curve intersects the x-axis (where $y = 0$).

Solve $x^2 + x - 12 = 0$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm \sqrt{49}}{2} = \frac{-1 \pm 7}{2}$ Thus, $x = 3$ and $x = -4$.

The definite integral from $x = -4$ to $x = 3$ will give us the area: $\text{Area} = \int_{-4}^{3} (x^2 + x - 12) \, dx$ Let's compute this integral step by step:

1. $\int x^2 \, dx = \frac{x^3}{3}$
2. $\int x \, dx = \frac{x^2}{2}$
3. $\int -12 \, dx = -12x$

Thus, the integral becomes: $\left[ \frac{x^3}{3} + \frac{x^2}{2} - 12x \right]_{-4}^{3}$

Now, plug in the limits: $= \left( \frac{3^3}{3} + \frac{3^2}{2} - 12(3) \right) - \left( \frac{(-4)^3}{3} + \frac{(-4)^2}{2} - 12(-4) \right)$ Simplifying both parts: $= \left( 9 + 4.5 - 36 \right) - \left( -\frac{64}{3} + 8 - (-48) \right)$ $= -22.5 - \left( -\frac{64}{3} + 8 + 48 \right)$ $= -22.5 + \left( \frac{64}{3} - 56 \right)$ The final area is the absolute value of the result.

Problem 2: $y = \sec^2 x$, bounded by the y-axis, x-axis, and $x = \frac{\pi}{4}$

We need to compute the area under the curve $y = \sec^2 x$ from $x = 0$ to $x = \frac{\pi}{4}$.

The integral is: $\text{Area} = \int_0^{\frac{\pi}{4}} \sec^2 x \, dx$

We know that $\int \sec^2 x \, dx = \tan x$, so: $\text{Area} = \left[ \tan x \right]_0^{\frac{\pi}{4}} = \tan\left( \frac{\pi}{4} \right) - \tan(0)$ $= 1 - 0 = 1$ Thus, the area is 1.

Problem 3: $y = 2\sqrt{x-1}$, bounded by the x-axis, $x = 5$, and $x = 17$

We are tasked with finding the area between the curve and the x-axis from $x = 5$ to $x = 17$.

The integral is: $\text{Area} = \int_5^{17} 2\sqrt{x-1} \, dx$

To solve this, let $u = x-1$, so $du = dx$. The new limits of integration are $u(5) = 4$ and $u(17) = 16$.

The integral becomes: $2 \int_4^{16} \sqrt{u} \, du = 2 \left[ \frac{2}{3} u^{3/2} \right]_4^{16}$ [ = \frac{4}{3} \left[ (16)^{3/2} - (4)^{3/2} \right] = \frac{4}{3} \left[ 64 - 8 \right] ] $= \frac{4}{3} \times 56 = \frac{224}{3}$ Thus, the area is $\frac{224}{3}$.

Problem 4: $y = x\sqrt{x} + 5$, bounded by the x-axis, $x = -1$, and $x = 4$

This is a more complex problem due to the negative boundary $x = -1$. Since the function $y = x\sqrt{x} + 5$ is not defined for negative $x$, we need to recheck the boundaries and reconsider the interval for integration.

For now, I'll need to clarify how you'd like to handle the undefined interval in this problem. Would you like to modify the interval for valid values?

Let me know if you'd like more details on any part or need further clarifications! Here are five related questions to expand on this:

1. Would you like to visualize the curves for these problems?
2. Do you want a more detailed breakdown of how the integrals are computed?
3. Are there any specific steps in the integration process you'd like me to revisit?
4. Would you like to explore how to handle areas when the curve dips below the x-axis?
5. Are there any further questions about how definite integrals calculate area?

Tip: Always check the limits of integration to ensure the function behaves well over the entire interval.