Solution

Actual integral:

$\int_{0}^{1} (x^3 - cx^2)\,dx = \left[\frac{x^4}{4} - \frac{cx^3}{3}\right]_0^1 = \frac{1}{4} - \frac{c}{3}$

Trapezoidal rule with one interval:

$T = \frac{1}{2}[f(0) + f(1)]$

$f(0) = 0$

$f(1) = 1 - c$

$T = \frac{1}{2}(1 - c)$

Since exact:

$\frac{1}{4} - \frac{c}{3} = \frac{1}{2}(1 - c)$

Solving:

$\frac{1}{4} - \frac{c}{3} = \frac{1}{2} - \frac{c}{2}$

Multiply by 12:

$3 - 4c = 6 - 6c$

$2c = 3$

$c = 1.5$