Area enclosed by two functions, curve–curve, and curve–line combinations.
Class 12 · Mathematics · CBSE & JEEWhen two curves \(y = f(x)\) and \(y = g(x)\) enclose a region, the area is the integral of the difference of the upper and lower function over the interval \([a, b]\).
The key: find where they intersect to get limits \(a\) and \(b\), then always subtract lower from upper.
\(\displaystyle\int_a^b f(x)\,dx\) = area under \(f\) above x-axis. Subtracting \(\displaystyle\int_a^b g(x)\,dx\) removes the area under \(g\). What remains is the area between the two curves.
Identify the two curves — \(y = f(x)\) and \(y = g(x)\).
Find intersection points — Solve \(f(x) = g(x)\). These give the limits \(a\) and \(b\).
Determine which is upper — Check by substituting a test value in \((a, b)\). Whichever gives a larger \(y\) is the upper curve.
Set up and integrate — \(A = \int_a^b [f(x) - g(x)]\,dx\).
Check sign — If the result is positive, you're done. If negative, you've swapped upper/lower; take absolute value.
This is the most tested combination in both CBSE and JEE. The parabola and line intersect at two points, forming a closed region.
Step 1 — Intersections: \(x^2 = x \Rightarrow x(x-1)=0 \Rightarrow x=0,\;1\)
Step 2 — Upper/Lower: At \(x=0.5\): \(y=x=0.5\), \(y=x^2=0.25\). So \(y=x\) is upper.
Step 3 — Integrate:
\(A = \displaystyle\int_0^1 (x - x^2)\,dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2} - \dfrac{1}{3} = \mathbf{\dfrac{1}{6}}\) sq. units
Intersections: \(x^2-2=x \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1,\;2\)
Upper: At \(x=0\): line gives 0, parabola gives \(-2\). So \(y=x\) is upper.
\(A = \displaystyle\int_{-1}^2 [x-(x^2-2)]\,dx = \int_{-1}^2(x-x^2+2)\,dx\)
\(= \left[\dfrac{x^2}{2}-\dfrac{x^3}{3}+2x\right]_{-1}^2 = \left(2-\dfrac{8}{3}+4\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}-2\right) = \mathbf{\dfrac{9}{2}}\) sq. units
Intersections: From \(x^2=4y \Rightarrow y = x^2/4\). Substitute into \(y^2=4x\):
\(\dfrac{x^4}{16}=4x \Rightarrow x^4 = 64x \Rightarrow x(x^3-64)=0 \Rightarrow x=0,\;4\). Points: \((0,0)\) and \((4,4)\).
Upper: On \([0,4]\): \(y=2\sqrt{x}\) (from \(y^2=4x\)) is upper; \(y=x^2/4\) is lower.
\(A = \displaystyle\int_0^4\left(2\sqrt{x}-\dfrac{x^2}{4}\right)dx = \left[\dfrac{4x^{3/2}}{3}-\dfrac{x^3}{12}\right]_0^4 = \dfrac{32}{3}-\dfrac{16}{3} = \mathbf{\dfrac{16}{3}}\) sq. units
Sometimes three or more boundaries (two lines and a curve, or three lines forming a triangle) define the region. Find all intersection points and set up accordingly.
Split at \(x = -1\) where \(|x+1|=0\).
For \(x\in[-4,-1]\): \(y=-(x+1)\). For \(x\in[-1,2]\): \(y=(x+1)\).
\(A = \displaystyle\int_{-4}^{-1}(-(x+1))\,dx + \int_{-1}^2(x+1)\,dx = \dfrac{9}{2}+\dfrac{9}{2} = \mathbf{9}\) sq. units
| Combination | Find Limits By | Formula | Typical Answer |
|---|---|---|---|
| \(y=x^2\) and \(y=x\) | Solve \(x^2=x\) | \(\int_0^1(x-x^2)dx\) | \(1/6\) |
| \(y^2=4x\) and \(x^2=4y\) | Solve simultaneously | \(\int_0^4(2\sqrt{x}-x^2/4)dx\) | \(16/3\) |
| \(y=x^2\) and \(y=4\) | \(x^2=4 \Rightarrow x=\pm2\) | \(\int_{-2}^2(4-x^2)dx\) | \(32/3\) |
| Circle & Parabola | Simultaneous equations | Split if needed | Use symmetry |