This is a definite integral and one solution is to apply the left and right Riemann approximations between the specified limits. Using dt=0.1 (that is, dividing the interval into 40 subdivisions), we get the two results:
RL=19.19 and RR=18.93. The average of these produces a closer approximation, that is, 19.06.
The method has been applied in Excel. It involved setting up a table of t values: -2, -1.9, -1.8, ..., 1.8, 1.9, 2. Calculating f(t) (the given expression f(t)=√((3t2-4)2+(2t-3)2)) and then multiplying by dt=0.1.
Excel was then used to sum all f(t)dt in the interval, so approximating the integral. RL uses the sum of the areas of rectangles where the height of the rectangle is f(t); RR uses the sum of the areas of rectangles with height=f(t+dt). In each case dt=0.1 is the width of the rectangle.