Step 1: Apply the trapezoidal rule with a single interval [0, 1].
For a single interval \([a, b]\), the trapezoidal rule for approximating the integral \( \int_a^b f(x) \, dx \) is given by:
\[
\int_a^b f(x) \, dx \approx \frac{b - a}{2} [f(a) + f(b)]
\]
In this case, \( a = 0 \), \( b = 1 \), and \( f(x) = x^3 - cx^2 \). So,
\[
\int_0^1 (x^3 - cx^2) \, dx \approx \frac{1 - 0}{2} [f(0) + f(1)]
\]
\[
\int_0^1 (x^3 - cx^2) \, dx \approx \frac{1}{2} [(0^3 - c(0)^2) + (1^3 - c(1)^2)]
\]
\[
\int_0^1 (x^3 - cx^2) \, dx \approx \frac{1}{2} [0 + (1 - c)]
\]
\[
\int_0^1 (x^3 - cx^2) \, dx \approx \frac{1 - c}{2}
\]
Step 2: Calculate the exact value of the integral.
\[
\int_0^1 (x^3 - cx^2) \, dx = \left[ \frac{x^4}{4} - \frac{cx^3}{3} \right]_0^1
\]
\[
\int_0^1 (x^3 - cx^2) \, dx = \left( \frac{1^4}{4} - \frac{c(1)^3}{3} \right) - \left( \frac{0^4}{4} - \frac{c(0)^3}{3} \right)
\]
\[
\int_0^1 (x^3 - cx^2) \, dx = \frac{1}{4} - \frac{c}{3} - 0
\]
\[
\int_0^1 (x^3 - cx^2) \, dx = \frac{1}{4} - \frac{c}{3}
\]
Step 3: Set the approximate value equal to the exact value since the trapezoidal rule is exact.
\[
\frac{1 - c}{2} = \frac{1}{4} - \frac{c}{3}
\]
Step 4: Solve for \( c \).
Multiply both sides by 12 to eliminate the fractions:
\[
12 \times \frac{1 - c}{2} = 12 \times \left( \frac{1}{4} - \frac{c}{3} \right)
\]
\[
6(1 - c) = 3(1) - 4(c)
\]
\[
6 - 6c = 3 - 4c
\]
Rearrange the terms to solve for \( c \):
\[
6 - 3 = 6c - 4c
\]
\[
3 = 2c
\]
\[
c = \frac{3}{2}
\]
Step 5: Select the correct answer.
The value of \( c \) is \( 3/2 \), which corresponds to option 2.