3.7 Q-5

Question Statement

Find the area between the $x$-axis and the curve $y = 4x - x^2$.

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Background and Explanation

To solve this problem, we need to calculate the area under the curve $y = 4x - x^2$ between the points where the curve intersects the $x$-axis. This is done by integrating the function over the interval where it is above the $x$-axis. The general approach involves setting up a definite integral and calculating the area bounded by the curve and the $x$-axis.

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Solution

Step 1: Express the function

The given function is:

$$y = 4x - x^2 = x(4 - x)$$

This is a quadratic function that cuts the $x$-axis at the points $x = 0$ and $x = 4$ (where $y = 0$).

Step 2: Identify the limits of integration

Since the curve is above the $x$-axis between $x = 0$ and $x = 4$, we will integrate over the interval $[0, 4]$.

Step 3: Set up the integral

We now set up the integral to find the area:

$$A = \int_{0}^{4} (4x - x^2) , dx$$

Step 4: Compute the integral

First, we integrate the terms $4x$ and $x^2$:

$$\int 4x , dx = 2x^2, \quad \int x^2 , dx = \frac{x^3}{3}$$

Now, substitute these results into the integral:

$$A = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4$$

Step 5: Evaluate the definite integral

Now, evaluate the expression at the limits $x = 4$ and $x = 0$:

$$A = \left( 2(4)^2 - \frac{(4)^3}{3} \right) - \left( 2(0)^2 - \frac{(0)^3}{3} \right)$$

Simplify the terms:

$$A = \left( 2(16) - \frac{64}{3} \right) - (0)$$ $$A = 32 - \frac{64}{3}$$

Step 6: Final simplification

Simplify the expression to get the final area:

$$A = \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$$

Thus, the area under the curve is $\frac{32}{3}$ 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:

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

• Power Rule of Integration:

$$\int x^n , dx = \frac{x^{n+1}}{n+1}$$

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Summary of Steps

1. Express the function: $y = 4x - x^2$
2. Set up the integral: $A = \int_0^4 (4x - x^2) , dx$
3. Integrate the function: $A = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4$
4. Evaluate the integral at $x = 4$ and $x = 0$: $A = 32 - \frac{64}{3}$
5. Simplify to get the final area: $A = \frac{32}{3} , \text{sq. units}$

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