The center of this ellipse is on the #x# axis, so the area of the ellipse is twice the area of the upper half of the ellipse.

Solve #x^2/5+y^2/3 = 1# for #y# to get the upper half of the ellipse.

The upper half of the ellipse is described by the function

#y = f(x) = sqrt((15-3x^2)/5)#.

Note that the values of #x# vary from #-sqrt5# to #sqrt5#.

To find the area using the trapezoidal rule we need an positive integer #n#. Since none is given we'll just go through the general method of solution.

#{a,b] = [-sqrt5, sqrt5]#, and

.

So we get

#Delta x = (b-a)/n = (sqrt5-(-sqrt5))/n = (2sqrt5)/n#

The endpoints of the subintervals are found by beginning at #a=-sqrt5# and successively adding #Delta x = (2sqrt5)/n# to find the points until we get to #x_n = b = pi#.

#x_0 = -sqrt5#, #x_1 = -sqrt5 + (2sqrt5)/n#, #x_2 = -sqrt5 + 2(2sqrt5)/n#, . . . , #x_i = -sqrt5 + i(2sqrt5)/n#, . . . , and #x_n = -sqrt5 + n(2sqrt5)/n= sqrt5 = b#

Now apply the formula (do the arithmetic) for #f(x) = sqrt((15-3x^2)/5)#..

#T_n=(Deltax)/2 [f(x_0)+2f(x_1)+2f(x_2)+ * * * +2f(x_(n-1))+f(x_n)] #

The total area of the ellipse is #~~2T_n#