Let T be the triangle with vertices `(0,0), (0,c^2 )and (c, c^2)` and let R be the region between `y=cx and y = x^2` where `c > 0` then

Options

A. Area `(R)=(c^3)/(6)`
B. Area of `R = (c^3)/(3)`
C. `overset(lim)(c to0^(+))(Area(T))/(Area(R))=3`
D. `overset(lim)(c to0^(+))(Area(T))/(Area(R))=3/2`

Answer

To solve the problem, we need to find the area of the triangle \( T \) with vertices at \( (0,0) \), \( (0,c^2) \), and \( (c,c^2) \), and the area of the region \( R \) between the lines \( y = cx \) and \( y = x^2 \) where \( c > 0 \).

### Step 1: Area of Triangle \( T \)

The area of a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]

For our triangle \( T \), the vertices are:
- \( (0, 0) \)
- \( (0, c^2) \)
- \( (c, c^2) \)

Plugging in the coordinates into the formula:

\[ \text{Area} = \frac{1}{2} \left| 0(c^2 - c^2) + 0(c^2 - 0) + c(0 - c^2) \right| \]

This simplifies to:

\[ \text{Area} = \frac{1}{2} \left| 0 + 0 - c^3 \right| = \frac{1}{2} \cdot c^3 = \frac{c^3}{2} \]

### Step 2: Area of Region \( R \)

Next, we need to find the area between the curves \( y = cx \) and \( y = x^2 \). First, we find the points of intersection by setting \( cx = x^2 \):

\[ x^2 - cx = 0 \]

Factoring out \( x \):

\[ x(x - c) = 0 \]

This gives us the points \( x = 0 \) and \( x = c \).

The area \( R \) between the curves from \( x = 0 \) to \( x = c \) can be calculated using the integral:

\[ \text{Area} = \int_0^c (cx - x^2) \, dx \]

Calculating the integral:

\[ \text{Area} = \int_0^c cx \, dx - \int_0^c x^2 \, dx \]

Calculating each integral separately:

1. For \( \int_0^c cx \, dx \):

\[ \int_0^c cx \, dx = \left[ \frac{cx^2}{2} \right]_0^c = \frac{c(c^2)}{2} = \frac{c^3}{2} \]

2. For \( \int_0^c x^2 \, dx \):

\[ \int_0^c x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^c = \frac{c^3}{3} \]

Now, substituting back into the area calculation:

\[ \text{Area} = \frac{c^3}{2} - \frac{c^3}{3} \]

To combine these fractions, we find a common denominator (which is 6):

\[ \text{Area} = \frac{3c^3}{6} - \frac{2c^3}{6} = \frac{c^3}{6} \]

### Final Result

Thus, the areas are:
- Area of triangle \( T \): \( \frac{c^3}{2} \)
- Area of region \( R \): \( \frac{c^3}{6} \)