The function given is f(x) = 36 - x2
We need to find the area bounded by this function, and the x axis, between x = -6 and x = 6. In other words, we need to find the definite integral between these x values, which is expressed as:
6
∫ (36 - x2) dx
-6
The function is a quadratic, which gives a parabola. In this case, it's inverted, with the vertex at (0, 36) and with roots (x intercepts) at -6 and 6 (±√36). It is symmetrical about the y axis.
A quick table will help you with the sketch:
x--0----1----2----3----4----5----6
y-36---35--32--27--20---11---0
We can sketch as required.
As it's symmetrical about the y axis, a quick approximation is the triangle between origin, x intercept, and vertex (y intercept). This is hw/2 and as there are 2 triangles, we need hw or 36.6 = 216 sq units. So this is my initial guesstimate (low, as triangle entirely inside curve) to check the accuracy of my later workings.
The midpoint rule is another way to approximate the area under the graph.
We have an interval of -6 to 6, or 12 units, and with n = 4 we will use rectangles with base of 3 units (12/4). As the function is symmetrical, we need to calculate the area (hb) of just 2 rectangles (0-3, 3-6) and then double.
the midpoints are 1.5 and 4.5
the y values (heights) are 36 - (1.5)2 = 33.75 and 36 - (4.5)2 = 15.75
the areas are 3 x 33.75 = 101.25 and 3 x 15.75 = 47.25
hence our approximation is 2 x (101.25 + 47.25) = 297 units2
For most accuracy, we must integrate the function.
The indefinite integral ∫ (36 - x2) dx = 36x - 1/3x3 + C
And therefore the definite integral between 6 and -6 is
[36x - 1/3x3 + C]6 - [36x - 1/3x3 + C]-6
=> ( 36(6) - 1/3(6)3 + C ) - ( 36(-6) - 1/3(-6)3 + C )
=> 216 - 72 + C + 216 - 72 - C
=> 216 + 216 - 72 - 72 = 288
Hence the area bounded by the function and x axis is 288 units2
We can plot on Desmos Graphing or similar, if you want to see what it should look like.
https://www.desmos.com/calculator/hgkdgorjkd
