Let's solve Question 39 and Question 43 by finding the area between the x-axis and the function over the specified intervals.
Question 39: f(x)=2−2x2 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:
2−2x2=0⟹2x2=2⟹x2=1⟹x=±1
Since the interval is [0,5], the function crosses the x-axis at x=1. So, we split the integral into two parts:
From x=0 to x=1, the function is above the x-axis.
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:
Area=∫01(2−2x2)dx+∫15∣2−2x2∣dx
Since the function is negative on [1,5], we integrate its opposite:
Area=∫01(2−2x2)dx−∫15(2−2x2)dx
Step 3: Compute the integrals.
Integral from 0 to 1:
∫01(2−2x2)dx=[2x−32x3]01=(2(1)−32(1)3)−(0)=2−32=34
Integral from 1 to 5:
∫15(2−2x2)dx=[2x−32x3]15
Evaluating the bounds:
= \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}$$
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### **Question 43: $$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!
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Math Problem Analysis
Mathematical Concepts
Definite Integrals Area under a Curve Splitting Integrals at x-intercepts Exponential Functions
Formulas
Area = ∫(a to b) |f(x)| dx Integral of e^x: ∫ e^x dx = e^x + C Integral of polynomial: ∫ x^n dx = (x^(n+1))/(n+1) + C