How to Find the Area Between Two Curves Using Integration
Master integration applications with this complete walkthrough of finding the area between curves. Learn how to find intersection points, set up the integral correctly, and avoid the common mistakes that cost students points.
📹 Video Walkthrough: This Exact Problem
Watch the full solution for finding the area between two curves step-by-step.
Table of Contents
The Problem
Find the area:
Find the area of the region bounded by the curves y = x² and y = 2x + 3.
Express your answer as an exact value.
This is one of those classic calculus problems that shows up everywhere. You've got two curves, and you need to find the area of the region they enclose. This is exactly what integration was made for.
The key steps are: find where the curves intersect, figure out which one is on top, then integrate the difference. If you're stuck on a similar problem or need to see the calculation worked out in detail, you can always generate a custom video solution on Torial.
Finding Intersection Points
First, we need to find where these two curves meet. That's where they intersect, and those intersection points will be our limits of integration.
Step 1: Set the equations equal
At the intersection points, both curves have the same y-value, so:
x² = 2x + 3
Step 2: Solve for x
x² = 2x + 3
x² - 2x - 3 = 0
Factor: (x - 3)(x + 1) = 0
So x = 3 or x = -1
Step 3: Find the y-coordinates
Plug these x-values into either equation (they'll give the same y):
When x = -1: y = (-1)² = 1 or y = 2(-1) + 3 = 1
When x = 3: y = (3)² = 9 or y = 2(3) + 3 = 9
Intersection points: (-1, 1) and (3, 9)
Integration limits:
x = -1 to x = 3
Good. Now we know the region goes from x = -1 to x = 3. Next, we need to figure out which curve is on top in this interval. Need more help with finding intersections? Check out other calculus videos in our library.
Setting Up the Integral
The area between two curves is found by integrating the difference between the top function and the bottom function. The formula is:
Area = ∫[a to b] (top - bottom) dx
Where a and b are the x-coordinates of the intersection points
Key Insight: Always subtract the bottom function from the top function. The result will be positive, giving you the area.
If you accidentally do bottom minus top, you'll get a negative answer. That's a red flag that you've got them backwards.
For our problem, we need to determine which function is on top between x = -1 and x = 3. Let's check a point in the middle, like x = 0.
Identifying Top and Bottom Functions
We need to figure out which curve is above the other in our interval. Let's test a point between x = -1 and x = 3.
Test at x = 0:
y = x²: y = 0² = 0
y = 2x + 3: y = 2(0) + 3 = 3
At x = 0, the line y = 2x + 3 is above the parabola y = x².
Verify at the endpoints:
At x = -1: both give y = 1 (they intersect)
At x = 3: both give y = 9 (they intersect)
Between the intersections, the line is always above the parabola.
Setup:
Top function: y = 2x + 3
Bottom function: y = x²
Area = ∫[-1 to 3] [(2x + 3) - x²] dx
Perfect. Now we can set up the integral. The top minus bottom gives us the height of a vertical slice, and integrating from -1 to 3 adds up all those slices to get the total area.
Performing the Integration
Time to integrate. First, simplify the integrand, then find the antiderivative.
Step 1: Simplify the integrand
Area = ∫[-1 to 3] [(2x + 3) - x²] dx
= ∫[-1 to 3] (2x + 3 - x²) dx
= ∫[-1 to 3] (-x² + 2x + 3) dx
Step 2: Find the antiderivative
Using the power rule for each term:
∫(-x² + 2x + 3) dx = -x³/3 + x² + 3x + C
We'll evaluate the definite integral, so we don't need the +C.
Step 3: Evaluate from -1 to 3
Using the Fundamental Theorem of Calculus:
Area = [-x³/3 + x² + 3x] evaluated from -1 to 3
= [-(3)³/3 + (3)² + 3(3)] - [-(-1)³/3 + (-1)² + 3(-1)]
= [-27/3 + 9 + 9] - [1/3 + 1 - 3]
= [-9 + 18] - [1/3 - 2]
= 9 - [-5/3]
= 9 + 5/3
= 27/3 + 5/3
= 32/3
That's the area. Let's make sure we didn't make any algebra mistakes by double-checking our work. Want to see more worked examples? Browse through hundreds of calculus solutions on Torial.
Simplifying the Result
Final Answer:
Area = 32/3 square units
Or approximately 10.67 square units
Double-check the calculation:
At x = 3: -27/3 + 9 + 9 = -9 + 18 = 9
At x = -1: -(-1)/3 + 1 - 3 = 1/3 + 1 - 3 = 1/3 - 2 = -5/3
Difference: 9 - (-5/3) = 9 + 5/3 = 32/3 ✓
The answer checks out. The area between the parabola and the line is 32/3 square units. This makes sense visually too: the region is a reasonable size, not too small and not impossibly large.
Interpreting the Answer
What does this answer actually mean?
Geometric meaning
The area of 32/3 square units represents the size of the region enclosed by the parabola y = x² and the line y = 2x + 3, between their intersection points at x = -1 and x = 3.
This is the actual two-dimensional area of that curved region on the coordinate plane.
Why integration works
We're essentially adding up infinitely many thin vertical rectangles. Each rectangle has width dx and height (top - bottom). Integrating sums all these rectangles to get the total area.
This is the geometric interpretation of the definite integral: area under a curve (or between curves).
Units
Since we're working with x and y coordinates (which are unitless), the area is in "square units." If the axes represented actual physical measurements, the units would be squared (like square meters or square feet).
Understanding what the answer represents helps you verify it makes sense. If you calculated an area of -5 or 1000 for this region, you'd know something went wrong. If you need help interpreting area calculations for different problems, you can generate a custom video solution for your specific question.
Common Mistakes to Avoid
Here's where students typically lose points. Learn these now so you don't make them on exam day. If you want personalized help avoiding these errors, create a custom study video for your specific problem.
❌ Mistake #1: Wrong integration limits
Using the y-coordinates of intersection points instead of x-coordinates, or using the wrong intersection points entirely.
Fix: Always use the x-coordinates of the intersection points as your limits. Since we're integrating with respect to x, the limits must be x-values.
❌ Mistake #2: Subtracting in the wrong order
Writing ∫(bottom - top) dx instead of ∫(top - bottom) dx. This gives a negative answer.
Fix: Always do top minus bottom. If you get a negative answer, you've got them backwards. Switch the order.
❌ Mistake #3: Not finding all intersection points
Missing an intersection point, or including points that aren't actually intersections. This gives the wrong region.
Fix: Always set the equations equal and solve completely. Check your solutions by plugging them back into both equations.
❌ Mistake #4: Assuming which function is on top
Guessing which curve is above the other without checking. Sometimes the curves cross, and the top function changes.
Fix: Always test a point in your interval to see which function gives a larger y-value. If they cross, you'll need to split the integral.
❌ Mistake #5: Algebra errors in integration
Making mistakes when finding antiderivatives or evaluating at the limits. Especially common with negative signs and fractions.
Fix: Write out every step. Double-check your power rule: ∫x^n dx = x^(n+1)/(n+1). Watch your signs when evaluating.
❌ Mistake #6: Forgetting to simplify the integrand
Trying to integrate (2x + 3) - x² without first combining terms. This makes the integration harder and more error-prone.
Fix: Always simplify the integrand first. Combine like terms, then integrate term by term.
Practice Problems with Video Solutions
Best way to get good at this? Practice. Try these similar problems and check your work with video solutions. You can also generate instant video explanations for any calculus problem you're working on.
Practice Problem 1: Different Curves
Find the area between y = x² - 4 and y = 2x.
Hint: Set x² - 4 = 2x, solve to get x = -2 and x = 4. Check which is on top. You should get Area = 36.
Get instant video solution on Torial →Practice Problem 2: Curves That Cross
Find the area between y = x³ and y = x from x = -1 to x = 1.
Hint: The curves cross at x = 0. You'll need to split the integral into two parts: from -1 to 0, and from 0 to 1. The top function switches.
Get instant video solution on Torial →Practice Problem 3: Horizontal Integration
Find the area between x = y² and x = y + 2.
Hint: This one requires integrating with respect to y instead of x. Set up: Area = ∫(right - left) dy. You should get Area = 9/2.
Get instant video solution on Torial →Practice Problem 4: Three Curves
Find the area of the region bounded by y = x², y = 2x, and y = x + 2.
Hint: Find all intersection points. The region has three boundaries. You might need to split this into two separate integrals.
Get instant video solution on Torial →Integrating with Respect to x vs. y
Should you integrate with respect to x or y?
✓ Integrate with Respect to x When:
- Functions are given as y = f(x)
- Vertical slices are easier to work with
- The top and bottom functions are clear
- Most common case
✓ Integrate with Respect to y When:
- Functions are given as x = f(y)
- Horizontal slices are easier
- You'd need multiple integrals with x
- Right and left boundaries are simpler
For this problem? Integrating with respect to x works perfectly since both functions are y = f(x) and the top/bottom is clear. When you're learning, it helps to have a step-by-step video walkthrough that shows exactly which method to use and when to switch.
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