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Let's solve **Question 39** and **Question 43** by finding the area between the x-axis and the function over the specified intervals.

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### **Question 39: \( f(x) = 2 - 2x^2 \) on the interval \([0, 5]\)**

#### Step 1: Check if the function crosses the x-axis.
We find the x-intercepts by setting \( f(x) = 0 \):
\[
2 - 2x^2 = 0 \implies 2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1
\]
Since the interval is \([0, 5]\), the function crosses the x-axis at \(x = 1\). So, we split the integral into two parts:  
1. From \(x = 0\) to \(x = 1\), the function is **above** the x-axis.  
2. From \(x = 1\) to \(x = 5\), the function is **below** the x-axis.

#### Step 2: Set up the integrals.
For the total area, we take the absolute value of each part:
\[
\text{Area} = \int_0^1 (2 - 2x^2) \, dx + \int_1^5 |2 - 2x^2| \, dx
\]
Since the function is negative on \([1, 5]\), we integrate its opposite:
\[
\text{Area} = \int_0^1 (2 - 2x^2) \, dx - \int_1^5 (2 - 2x^2) \, dx
\]

#### Step 3: Compute the integrals.

1. **Integral from 0 to 1**:
\[
\int_0^1 (2 - 2x^2) \, dx = \left[ 2x - \frac{2x^3}{3} \right]_0^1 = \left( 2(1) - \frac{2(1)^3}{3} \right) - (0) = 2 - \frac{2}{3} = \frac{4}{3}
\]

2. **Integral from 1 to 5**:
\[
\int_1^5 (2 - 2x^2) \, dx = \left[ 2x - \frac{2x^3}{3} \right]_1^5
\]
Evaluating the bounds:
\[
\left( 2(5) - \frac{2(5)^3}{3} \right) - \left( 2(1) - \frac{2(1)^3}{3} \right)
= \left( 10 - \frac{250}{3} \right) - \left( 2 - \frac{2}{3} \right)
\]
\[
= \left( 10 - 83.33 \right) - \left( 2 - 0.67 \right) = -73.33 + 1.33 = -72
\]
Since this area is negative, we take the absolute value: \(72\).

#### Step 4: Total Area.
\[
\text{Total Area} = \frac{4}{3} + 72 = \frac{4}{3} + \frac{216}{3} = \frac{220}{3}
\]

---

### **Question 43: \( f(x) = e^x - 1 \) on the interval \([-1, 2]\)**

#### Step 1: Check if the function crosses the x-axis.
Set \( f(x) = 0 \) to find where the graph intersects the x-axis:
\[
e^x - 1 = 0 \implies e^x = 1 \implies x = 0
\]
So, the function crosses the x-axis at \(x = 0\). We need to split the integral:
1. From \(-1\) to \(0\), the function is **below** the x-axis.
2. From \(0\) to \(2\), the function is **above** the x-axis.

#### Step 2: Set up the integrals.
\[
\text{Area} = \int_{-1}^0 |e^x - 1| \, dx + \int_0^2 (e^x - 1) \, dx
\]
Since the function is negative on \([-1, 0]\), we integrate the opposite:
\[
\text{Area} = -\int_{-1}^0 (e^x - 1) \, dx + \int_0^2 (e^x - 1) \, dx
\]

#### Step 3: Compute the integrals.

1. **Integral from \(-1\) to \(0\)**:
\[
\int_{-1}^0 (1 - e^x) \, dx = \left[ x - e^x \right]_{-1}^0
\]
Evaluating the bounds:
\[
(0 - e^0) - \left( -1 - e^{-1} \right) = -1 + 1 + \frac{1}{e} = \frac{1}{e}
\]

2. **Integral from \(0\) to \(2\)**:
\[
\int_0^2 (e^x - 1) \, dx = \left[ e^x - x \right]_0^2
\]
Evaluating the bounds:
\[
\left( e^2 - 2 \right) - \left( e^0 - 0 \right) = e^2 - 2 - 1 = e^2 - 3
\]

#### Step 4: Total Area.
\[
\text{Total Area} = \frac{1}{e} + (e^2 - 3)
\]

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### **Final Answers**

1. **Question 39:** The total area is:
   \[
   \frac{220}{3}
   \]

2. **Question 43:** The total area is:
   \[
   \frac{1}{e} + (e^2 - 3)
   \]

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Would you like further steps or clarifications?

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**Related Questions:**
1. What is the difference between positive and negative integrals?
2. How do you handle absolute values in definite integrals?
3. How does splitting the integral at x-intercepts affect the calculation?
4. What is the geometric meaning of a negative integral?
5. How do exponential functions behave over positive and negative intervals?

**Tip:** Always check where the graph crosses the x-axis to ensure correct area calculations!

Detailed solution for finding the area between the x-axis and functions using definite integrals for questions 39 and 43. Explore how to handle x-intercepts, absolute values in integrals, and areas under curves for quadratic and exponential functions. Suitable for students in Grades 11-12.

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