How to Find the Area of the Shaded Region the Graph Depicts

To find the area of the shaded region the graph depicts, identify the region’s boundaries (the curves above and below and the left and right limits), then set up a definite integral of the upper boundary minus the lower boundary, and evaluate it. The result is always a positive number. For example, the area under the line y = x from x = 0 to x = 2 is the integral of x dx over [0, 2], which equals ½(2)² − ½(0)² = 2 square units. That single idea — integrate (top function minus bottom function) over the stated interval — solves nearly every shaded-region problem you’ll meet in calculus.

Identify the Shaded Region

The shaded region is the specific patch of the coordinate plane enclosed by the graph’s curves that the problem asks you to measure. Your first job is to pin down exactly which patch that is. Shading usually signals it visually, but you still need to read the boundaries off the axes.

Most shaded-region problems fall into one of three shapes:

• Area under a single curve: the region between one function and the x-axis, bounded on the left and right by vertical lines (or points where the curve crosses the axis).
• Area between two curves: the region trapped between an upper function and a lower function over a shared x-interval. The lower boundary might be the x-axis or a second curve.
• Inner or outer regions: with circles or closed loops, “inner” means the smaller enclosed area and “outer” means the space between an outer boundary and an inner one.

Find the x-values where the shaded area starts and stops — these are your integration limits. Then note the curve or line sitting on top and the one on the bottom throughout the interval. If the top and bottom swap places inside the interval, you’ll need to split the integral at the crossing point. Khan Academy’s treatment of area between curves walks through this boundary-reading step with interactive graphs, which is a good way to build the habit of scanning left-to-right.

Set Up the Integral

A definite integral adds up infinitely thin rectangles to total the area. To translate a shaded region into one, write “upper boundary minus lower boundary” as the integrand and place the graph’s x-limits on the integral sign.

The general form is:

Area = ∫ from a to b of [f(x) − g(x)] dx, where f(x) is the top curve, g(x) is the bottom curve, and a and b are the left and right limits.

This works fine even when the bottom curve is the x-axis, because then g(x) = 0 and the formula reduces to ∫ f(x) dx.

If the region is easier to describe left-to-right — every vertical line through it hits the same top and bottom curves — integrate with respect to x. If it’s easier top-to-bottom, or the curves are given as x in terms of y, integrate with respect to y instead, with “right minus left” as the integrand. Sometimes swapping the variable saves you from splitting one awkward integral into two. Sketch first, decide second.

Getting the order right (upper minus lower) matters more than anything else at this stage — flipping it gives you a negative number you’ll have to correct later.

Compute the Area

Evaluate the integral using an antiderivative, then apply the Fundamental Theorem of Calculus: plug in the upper limit, subtract the value at the lower limit. The NIST Engineering Statistics Handbook describes the same accumulate-and-evaluate logic in its treatment of continuous distributions, where area under a curve represents probability.

Suppose the graph shades the region between y = x and y = x² for 0 ≤ x ≤ 1. The line y = x is above the parabola here, so the integral is ∫ from 0 to 1 of (x − x²) dx. The antiderivative of x − x² is x²/2 − x³/3. Evaluating at 1 gives 1/2 − 1/3 = 1/6. Evaluating at 0 gives 0. So the area is 1/6 − 0 = 1/6 square units.

Handling curves below the axis

The integral mechanically computes signed area: regions below the x-axis contribute negative values. Real geometric area is never negative, so take the absolute value of any piece that lies under the axis. If part of the region is above and part below, split it at the x-intercepts and add the absolute values of each piece. This is the single most common place where students lose points on shaded-region problems — the setup looks right, but a negative contribution gets carried through uncorrected.

Check and Interpret

Before writing your final answer, sanity-check it. A few quick habits catch nearly all mistakes.

• Re-sketch. Redraw the shaded region and ask whether your numeric answer seems plausible relative to the grid. A region roughly 2 units wide and 3 tall should have an area near 6, not 600.
• Use symmetry. If the region is symmetric about the y-axis, the area on one side doubled should equal the total. If your answer violates a symmetry the graph clearly has, something is wrong.
• Confirm positivity. Area is always a non-negative quantity. A negative answer means either the boundaries were reversed or the integrand needs an absolute value.
• Check the units. If the axes carry units — meters, seconds, dollars — the area carries the square of those units. Report them.

If your result survives all four checks, you’ve found the area of the shaded region with confidence.

Work Through a Between-Curves Example Step by Step

Consider a graph that shades the region between y = 4 − x² and the x-axis, from x = −2 to x = 2. Here the parabola sits entirely above the axis across the interval, so the lower boundary is simply y = 0.

Set up the integral as ∫ from −2 to 2 of (4 − x²) dx. The antiderivative of 4 − x² is 4x − x³/3. Evaluating at the upper limit of 2 gives 8 − 8/3 = 16/3. Evaluating at the lower limit of −2 gives −8 + 8/3 = −16/3. Subtracting gives 16/3 − (−16/3) = 32/3, about 10.67 square units.

Notice how the symmetry paid off here: because the parabola is even and the interval is centered on zero, you could just as easily have computed 2 × ∫ from 0 to 2 of (4 − x²) dx and reached the same 32/3. Recognizing symmetry early often halves your arithmetic load.

When the top and bottom curves swap, the setup changes. If two curves cross inside the interval, test a sample point to see which sits on top in each piece, then split the integral at the crossing and keep the order correct for each part. Blindly applying one integrand across the whole span silently mixes positive and negative contributions and gives a wrong answer.

Key Takeaways

• Area of a shaded region = definite integral of (upper boundary − lower boundary) between the graph’s left and right limits.
• Read the boundaries off the graph first: identify the top curve, bottom curve, and the x-values where the region begins and ends.
• The Fundamental Theorem of Calculus finishes the job: evaluate an antiderivative at the upper limit and subtract its value at the lower limit.
• Areas below the x-axis come out negative from the integral — take absolute values so the final answer is positive.
• Verify with a quick sketch, symmetry, and unit check; a geometric area must always be a positive number in squared units.