Use the Midpoint Rule with n = 4 to approximate the area of the region bounded by the graph of f and the x-axis over the interval. Function Interval f(x) = 2x2 [0, 2]

Question

Use the Midpoint Rule with n = 4 to approximate the area of the region bounded by the graph of f and the x-axis over the interval.

FunctionInterval
f(x) = 2x2	[0, 2]

Answer 1

y = 2x² is a quadratic equation and the graph is a parabola.

x = 0, y = 0

x = 1, y = 2

x = 2, y = 8

So the vertex of the parabola is at the origin (0,0) and it is symmetrical about the y axis. The parabola passes through (2,8) and (-2,8).

We need to find the area between the graph and the x axis, between x = 0 and x = 2.

This area is the definite integral, written as:

₂

∫ (2x²) dx

⁰

Between (0,0) and (2,8) we can approximate the curve with a triangle with base 2 and height 8. The area is 1/2 hb = 8 units². So this is my quick (bit over) guesstimate.

To calculate the area accurately, we would need to integrate the function.

The indefinite integral ∫ (2x²) dx = ²/₃x³ + C

and the definite integral is [²/₃x³ + C]² - [²/₃x³ + C]⁰

The Midpoint Rule gives us a way to approximate this (using areas of rectangles).

The interval required is 0 to 2, or 2 units. With n = 4, we have 4 rectangles with a base of 1/2.

The bases of the rectangles are 0 to 1/2, 1/2 to 1, 1 to 1 1/2, 1 1/2 to 2.

Next, the midpoints are 1/4. 3/4, 5/4 and 7/4

And then we need the y values (= heights)

2(1/4)² = 1/8

2(3/4)² = 9/8

2(5/4)² = 25/8

2(7/4)² = 49/8

And now we can calculate the area of each rectangle hb:

1/8.1/2 = 1/16

9/8.1/2 = 9/16

25/8.1/2 = 25/16

49/8.1/2 = 49/16

Total area = 84/16 = 42/8 = 21/4 = 5.25 units²

Evaluating the definite integral (given above) gives ²/₃(2)³ - ²/₃(0)³ = ¹⁶/₃ - 0 = 5 ¹/₃, so in this case the Midpoint Rule has given a very close approximation.

You can use Desmos Graphing or similar to see what it looks like.

https://www.desmos.com/calculator/1oq6ndtr7s