Question 1028629
1. *[illustration cx1.JPG].
2. Area of the trapezoid marked by the black dashed boundaries and the graph.
3. You could just calculate the area of the trapezoid or using a {{{dx}}}, make a summation.
{{{A=sum(P(x)*dx,x=1,N)}}}
4. {{{int(P(t),dt,5,15)=100(t^(1.4+1))/(2.4)+3t^2+C}}}
{{{int(P(t),dt,5,15)=(125/3)(15^(2.4)-5^(2.4))+3(15^2-5^2)}}}
{{{int(P(t),dt,5,15)=(125/3)(664.69-47.59)+3(225-25)}}}
{{{int(P(t),dt,5,15)=(125/3)(617.1)+3(200)}}}
{{{int(P(t),dt,5,15)=25712.5+600}}}
{{{int(P(t),dt,5,15)=26312.5}}}{{{Watt-s}}}or {{{J}}}
As a comparison, the area of the trapezoid would yield,
{{{A=(1/2)(10)(4431.265+951.827)}}}
{{{A=27515.5}}}
Also, a summation using {{{dx=0.5}}} would yield,
{{{A=27061.3}}}
Even closer.