(Question 15) Show all work, thank you!answers (a) =67ft and (b) = 67.75ft **15. Approximating Displacement**The velocity in feet per second (ft/s) of an object moving along a line is given by the equation \( v = 3t^2 + 1 \) on the interval \( 0 \leq t \leq 4 \), where \( t \) is measured in seconds.a. Divide the interval \([0, 4]\) into \( n = 4 \) subintervals, \([0, 1]\), \([1, 2]\), \([2, 3]\), and \([3, 4]\). On each subinterval, assume the object moves at a constant velocity equal to \( v \) evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on \([0, 4]\) (see part (a) of the figure).b. Repeat part (a) for \( n = 8 \) subintervals (see part (b) of the figure).**Explanation of the Figures:**- **Figure (a):** Shows the velocity function \( v = 3t^2 + 1 \) plotted as a curve. The interval \([0, 4]\) is divided into \( n = 4 \) subintervals. For each subinterval, a rectangle is drawn with its height equal to the velocity at the midpoint of the subinterval, representing the approximation of displacement. The areas of the rectangles provide an estimate of the total displacement.- **Figure (b):** Also shows the velocity function \( v = 3t^2 + 1 \), but here the interval \([0, 4]\) is divided into \( n = 8 \) subintervals. Similarly, rectangles denote the constant velocity assumption over each subinterval. With more subintervals, this approximation is closer to the actual displacement, as evidenced by the increased number of rectangles fitting more closely to the curve.

(a) sub-intervalsWidth of sub interval, Subintervals are: The midpoints of the subinterval are:Now evaluate the velocity at the midpoint: Midpoint, 0.51.751.57.752.519.753.537.75Total67Estimate the displacement using the midpoint rule:(b) sub intervalsWidth of sub interval, Subintervals are:The midpoints of the sub interval are:Now evaluate the velocity at the midpoint:Midpoint, 0.251.18750.752.68751.255.68751.7510.18752.2516.18752.7523.68753.2532.68753.7543.1875Total135.5Estimate the displacement using the midpoint rule:(a) 67 ft(b) 67.75 ft