Understanding the Problem: Finding Area Between Parabola and Line

The question asks us to calculate the area enclosed, or "cut off", between two curves: a parabola and a straight line. This is a classic problem in calculus that involves finding the intersection points of the two curves and then integrating the difference of their equations over the interval defined by these intersection points. Finding the Area between Parabola and Line requires careful steps.

Identifying the Equations

We are given the following equations:

• Parabola: \(4y = 3x^2\)
• Straight Line: \(2y = 3x + 12\)

To make it easier to work with, let's express both equations in terms of \(y\):

• Parabola: \(y = \frac{3}{4}x^2\)
• Straight Line: \(y = \frac{3}{2}x + 6\)

These are the specific equation forms we will use.

Finding the Intersection Points

The points where the parabola and the straight line intersect are crucial because they define the limits of our integration. To find these points, we set the \(y\) values from both equations equal to each other:

\(\frac{3}{4}x^2 = \frac{3}{2}x + 6\)

To eliminate fractions, multiply the entire equation by 4:

\(3x^2 = 6x + 24\)

Rearrange the terms to form a quadratic equation:

\(3x^2 - 6x - 24 = 0\)

Divide by 3 to simplify:

\(x^2 - 2x - 8 = 0\)

Now, we solve this quadratic equation for \(x\). We can factor it:

\((x - 4)(x + 2) = 0\)

This gives us two solutions for \(x\):

• \(x - 4 = 0 \implies x = 4\)
• \(x + 2 = 0 \implies x = -2\)

These \(x\) values, -2 and 4, are the limits of integration for finding the Area between Parabola and Line.

Setting up the Definite Integral for Area Calculation

To find the area between the two curves, we need to integrate the difference between the upper curve and the lower curve with respect to \(x\) from the lower limit (\(-2\)) to the upper limit (\(4\)).

In the interval \([-2, 4]\), the straight line \(y = \frac{3}{2}x + 6\) is above the parabola \(y = \frac{3}{4}x^2\). You can test a point in the interval, say \(x=0\). For the line, \(y = 6\). For the parabola, \(y = 0\). Since 6 > 0, the line is indeed above the parabola in this interval.

The area \(A\) is given by the definite integral:

\(A = \int_{-2}^{4} \left( y_{\text{line}} - y_{\text{parabola}} \right) dx\)

\(A = \int_{-2}^{4} \left( \left(\frac{3}{2}x + 6\right) - \left(\frac{3}{4}x^2\right) \right) dx\)

\(A = \int_{-2}^{4} \left( \frac{3}{2}x + 6 - \frac{3}{4}x^2 \right) dx\)

This integral calculation is a core concept in calculus for finding areas.

Evaluating the Definite Integral

Now, we evaluate the integral:

\(A = \left[ \frac{3}{2} \cdot \frac{x^2}{2} + 6x - \frac{3}{4} \cdot \frac{x^3}{3} \right]_{-2}^{4}\)

\(A = \left[ \frac{3}{4}x^2 + 6x - \frac{1}{4}x^3 \right]_{-2}^{4}\)

Now, we substitute the upper limit (4) and the lower limit (-2) into the expression and subtract:

\(A = \left( \frac{3}{4}(4)^2 + 6(4) - \frac{1}{4}(4)^3 \right) - \left( \frac{3}{4}(-2)^2 + 6(-2) - \frac{1}{4}(-2)^3 \right)\)

\(A = \left( \frac{3}{4}(16) + 24 - \frac{1}{4}(64) \right) - \left( \frac{3}{4}(4) - 12 - \frac{1}{4}(-8) \right)\)

\(A = \left( 12 + 24 - 16 \right) - \left( 3 - 12 + 2 \right)\)

\(A = \left( 36 - 16 \right) - \left( -9 + 2 \right)\)

\(A = 20 - (-7)\)

\(A = 20 + 7\)

\(A = 27\)

The area calculated using calculus methods is 27 square units. This demonstrates how to find the Area between Parabola and Line.

Conclusion

The area cut off the parabola \(4y = 3x^2\) by the straight line \(2y = 3x + 12\) is 27 square units. This confirms the result obtained through the integration process, a fundamental tool in calculus for finding the Area between Parabola and Line.