3.7 Q-2

Question Statement

Find the area above the $x$-axis and under the curve $y = 5 - x^2$ from $x = -1$ to $x = 2$.

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Background and Explanation

To solve this problem, we will use definite integration to calculate the area under the curve $y = 5 - x^2$ between the limits $x = -1$ and $x = 2$.

The curve intersects the $x$-axis at the points where $y = 0$, which can be found by solving $5 - x^2 = 0$. These points are at $x = -\sqrt{5}$ and $x = \sqrt{5}$.

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Solution

We need to compute the following integral to find the area:

$$A = \int_{-1}^{2} (5 - x^2) , dx$$

Step 1: Break the integral into two parts

The integral can be split into:

$$A = \int_{-1}^{2} 5 , dx - \int_{-1}^{2} x^2 , dx$$

Step 2: Evaluate the first integral

The integral of 5 is:

$$\int 5 , dx = 5x$$

Now evaluate this from $x = -1$ to $x = 2$:

$$\left[ 5x \right]_{-1}^{2} = 5(2) - 5(-1) = 10 + 5 = 15$$

Step 3: Evaluate the second integral

The integral of $x^2$ is:

$$\int x^2 , dx = \frac{x^3}{3}$$

Now evaluate this from $x = -1$ to $x = 2$:

$$\left[ \frac{x^3}{3} \right]_{-1}^{2} = \frac{(2)^3}{3} - \frac{(-1)^3}{3} = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3$$

Step 4: Combine the results

Now, subtract the results from the two integrals:

$$A = 15 - 3 = 12$$

Thus, the area under the curve is 12 square units.

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Key Formulas or Methods Used

• Definite Integral: The area under a curve from $x = a$ to $x = b$ is given by:

$$A = \int_{a}^{b} f(x) , dx$$

• Basic Integration Rules:
  • $\int a , dx = ax$
  • $\int x^2 , dx = \frac{x^3}{3}$

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Summary of Steps

1. Set up the integral: $A = \int_{-1}^{2} (5 - x^2) , dx$
2. Break the integral into two parts: $A = \int_{-1}^{2} 5 , dx - \int_{-1}^{2} x^2 , dx$
3. Evaluate the first integral: $\int 5 , dx = 5x \quad \Rightarrow , 15$
4. Evaluate the second integral: $\int x^2 , dx = \frac{x^3}{3} \quad \Rightarrow , 3$
5. Subtract the results: $A = 15 - 3 = 12 , \text{square units}$