Find the area of the region bounded by the curves y = x2 − 1 and y = cos(x). Give your answer correct to 2 decimal places.

Question

Find the area of the region bounded by the curves y = x² − 1 and y = cos(x). Give your answer correct to 2 decimal places.

Answer 1

It is always helpful to know the general shape of the two curves to start. In this case,

1. y=x²-1 is a concave up parabola, shifting the parent parabola, y=x² down by 1, crossing the y axis at y=-1 (solving for y after setting x=0) and crossing the x axis at ±1 (solving for x after setting y=0) and y being below the x axis (negative) for -1<x<+1 and above the x axis (positive) for x<-1 and x>1;

and

1. y=cos(x) crosses the y axis at y=1 (when x=0) and crosses the x axis at ±π/2 (when y=0) with the curve being concave down between x=-π/2 and x=π/2 with y being above the x axis (positive) for -π/2≤x≤+π/2 and below the x axis (negative) for x<-π/2 and x>π/2.

and since -π/2<-1 and π/2>1

1. y=x²-1 intersects with y=cos(x) somewhere above the x axis.

and

1. y=cos(x) will be above y=x²-1 between the intersection points, obvious since y=cos(x) is concave down and y=x²-1 is concave up between the points of intersection.

AND since, except at key points, it is usually difficult to find the exact intersection between a polynomial and a transcendental function, the exact intersection points of the two functions will be determined by graphing with a graphing device (e.g. a graphing calculator) rather than algebraically.

The next step is to subtract the area under the curve of the function with the lesser y values, y=x²-1, from the area under the curve of the function with the greater y values, y=cos(x), between the intersection points (the greater function being above the lesser function in the graph). This is similar to having to find the distance between two values on a number line by algebraically subtracting the lesser number from the greater number. Keep in mind that when the area of either of the functions are negative between the intersection points,

y=x²-1 in this case, the area is negative so that subtracting ends up being the same as adding the area's absolute value (a positive value), just as, on a number line, subtracting a negative number is the same as adding a positive number.

Because the NET area under a curve and the x axis between two x values is the integral of the curve's function between the two x values of the intersection points, we have

[Integral of cos(x)dx from the lesser x value of the intersecting points to the greater]

minus

[Integral (x²-1)dx from the lesser x value of the intersecting points to the greater]

where

The integral of (x²-1)dx is x³/3-x+C₁ and

The integral of cos(x)dx is sin(x)+C₂

where C₁ and C₂ are constants.

The END