Trapezoidal Rule

Consider f(t)=t^2. Using the trapezoidal rule with Δt = 0.25, find the area under the curve for 0≤t≤1.

Hint
$$$\int_{a}^{b}f(x)dx\approx T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdot \cdot \cdot +2f(x_{n-1})+f(x_n)]$$$
where $$\Delta x=\frac{b-a}{n}$$
Similar to solving a definite integral, the Trapezoidal Rule is used to find the area under a curve by dividing the total area into little trapezoids.
$$$\int_{a}^{b}f(x)dx\approx T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdot \cdot \cdot +2f(x_{n-1})+f(x_n)]$$$
where $$\Delta x=\frac{b-a}{n}$$
$$$\Delta x=\frac{b-a}{n}=\frac{1-0}{4}=0.25$$$
$$$Area=\frac{0.25}{2}[0^2+2(0.25)^2+2(0.5)^2+2(0.75)^2+(1)^2]$$$
$$$Area=0.344$$$
0.344