8.4 Finding the Area Between Curves Expressed as Functions of x

Learning objectives

—CHA-5.A—Calculate areas in the plane using the definite integral• CHA-5.A Calculate areas in the plane using the definite integral.• CHA-5.A.1 Areas of regions in the plane can be calculated with definite integrals.• Enduring understanding CHA-5: Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

When curves are written as functions of $x$ , use a vertical slice of width $dx$ . If $y=u(x)$ is above $y=\ell(x)$ from $x=a$ to $x=b$ , the slice height is $u(x)-\ell(x)$ , so integrating those rectangle areas gives the region’s area.

A=\int_a^b\big(\text{upper}(x)-\text{lower}(x)\big),dx

Find the left and right boundaries, often by solving the curve-intersection equation.
Test the interval to identify the upper and lower functions.
Write upper minus lower.
Split the integral at any point where their order changes.
Evaluate and report square units.

The curves $y=2x$ and $y=x^2$ intersect where $x^2=2x$ , so $x=0$ and $x=2$ . On $(0,2)$ , $2x>x^2$ . Therefore $A=\int_0^2(2x-x^2)\,dx=[x^2-x^3/3]_0^2=4-8/3=4/3$ square units.

A negative integral signals that the curves were subtracted in the wrong order on part or all of the interval; geometric area is never negative. Do not use one upper-minus-lower expression across an intersection where the curve order switches—split there and write a nonnegative slice height on each piece.

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