Let fx the integral from 2 to x of cos t24dt Using linear a

Let f(x) = the integral from 2 to x of cos (t^2-4)dt. Using linear approximaion, determine the approximate value of f at 1.98

We can see at x = 2, f(2) = 0 f\'(X) = cos(2x^2 - 4) f\'(1.98) = cos(2*1.98^2 - 4) = -0.765 So slope = -0.765 and passes through (2,,0) y - 0 = -0.765(x-2) At x= 1.98 y = -0.765(1.98-2) = 0.0153