Suppose $$f(t)=t^3$$, the area under the curve for $$0<=t<=4$$, estimated by using the trapezoidal rule with $$Deltat=0.5$$, is most nearly:

The trapezoidal rule is used when we want to estimate a definite integral. It essentially creates multiple trapezoids under the curve of f(t) spanning across our specified bounds. We define a subinterval length, $\Delta t$, that defines the x-axis length of each trapezoid.  

For our problem, if we know that the curve spans from  $0\le t\le 4$ and our $\Delta t$ is equal to 0.5, we have enough information to figure out how many unique trapezoids, n, we will create under our curve to estimate the definite integral:

$$\begin{gathered} \frac{b-a}{\Delta t}=n \\ \frac{4-0}{0.5}=n \\ 8 = n \end{gathered}$$

Therefore, we have 8 unique trapezoids. The equation for the trapezoidal rule with equally spaced base points is as follows:

${\int }_a^{b}f\left(x\right)dx\approx ∆x\left.open\frac{1}{2}\left[f\right(x_0)+f(x_n\left)\right]+\sum _{i=1}^{n-1}f\left(x_i\right)\right]$

Plugging in what we know at this point:

$$\begin{gathered} {\int }_0^{4}f\left(t\right)dt\approx ∆t\left.open\frac{1}{2}\left[f\right(t_0)+f(t_8\left)\right]+\sum _{i=1}^{8-1}f\left(t_i\right)\right] \\ {\int }_0^4{t}^{3}dt\approx (0.5)\left.open\frac{1}{2}\left[f\right(t_0)+f(t_8\left)\right]+\sum _{i=1}^{8-1}f\left(t_i\right)\right] \end{gathered}$$

We can all of our based points by starting at a (t = 0), adding our delta t value (0.5), until we get to b (t = 4):

$\{t_0,t_1,t_2,t_3,t_4,t_5,t_6,t_7,t_8\}=\{0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4\}$

Therefore:

$$\begin{gathered} {\int }_0^4{t}^{3}dt\approx (0.5)\left.open\frac{1}{2}\left[f\right(0)+f(4\left)\right]+\left[f\right(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5\left)\right]\right] \\ {\int }_0^4{t}^{3}dt\approx (0.5)\left.open\frac{1}{2}\left[\right(0^3)+(4^3\left)\right]+\left[\right(0.5^3)+(1^3)+(1.5^3)+(2^3)+(2.5^3)+(3^3)+(3.5^3\left)\right]\right] \\ {\int }_0^4{t}^{3}dt\approx (0.5)\left.open\frac{1}{2}(0+64)+(.125+1+3.375+8+15.625+27+42.875)\right] \\ {\int }_0^4{t}^{3}dt\approx (0.5)(32+98) \\ {\int }_0^4{t}^{3}dt\approx (0.5)(32+98) \\ {\int }_0^4{t}^{3}dt\approx 65 \end{gathered}$$

Thus, the area under the curve is most nearly 65.