Q.Find the area of the region bounded by the parabola y=x2 and the line y=4.
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Concept understanding — Area between Two Curves
When two curves y=f(x) and y=g(x) cross each other, they enclose a bounded region between them. To find its area, the first step is always to locate the points where the two curves meet, by solving their equations simultaneously -- these intersection points become the limits of integration, and they are almost never given directly in the problem statement.
Once the limits x=a and x=b are known, the enclosed area is the difference between the area under the upper curve and the area under the lower curve, over that same interval, taken as a positive quantity:
A=∫abf(x)dx−∫abg(x)dx
In practice this means identifying which of the two curves lies above the other throughout the interval (usually by testing a convenient point between the intersection points), subtracting the lower function from the upper one, and integrating that difference directly -- rather than integrating each curve separately against the X-axis and subtracting afterwards, which is equivalent but usually more work.
This idea covers a wide range of problems that look different on the surface but share the same method: the region between two parabolas, the region between a parabola and a straight line, the segment of a circle or an ellipse cut off by a chord, and the sector of a circle cut off by a line through the centre (where the region is naturally split into a triangular piece and a curved piece, each found by a different technique, and then added together). Sketching both curves and marking their intersection points before integrating is the single most reliable way to avoid setting up the wrong difference of integrals.
Find where y=x2 meets y=4, then integrate (line − parabola).
✓Final answer
The area is 332 square units.
The curves meet at x=±2; the horizontal line y=4 lies above the parabola between them.
Setting x2=4 gives x=±2. Testing x=0: the line gives y=4 and the parabola gives y=0, so y=4 lies above y=x2 throughout (−2,2). By the even symmetry of 4−x2, the area is
Solve x2=4 for the intersection points, confirm the line is the upper curve, and use the region's even symmetry to integrate over [0,2] and double, or integrate directly over [−2,2].
Integrating only over [0,2] without doubling for the symmetric left half; sign errors expanding 4x−x3/3 at x=2.