The area of the region enclosed by the parabolas y=x²-5x and y=7x-x² is ________.

Solution

Maths sir · 3 steps

Subtract the two parabolas first. The difference is one quadratic, and its area is quick to find.

1. Find the meeting points. Set x² - 5x = 7x - x². This gives 2x² - 12x = 0, so 2x(x - 6) = 0, and the curves meet at x = 0 and x = 6.
2. Pick the upper curve. Test x = 3. The curve y = 7x - x² gives 12 and y = x² - 5x gives -6. So y = 7x - x² is on top, and the gap between the curves is (7x - x²) - (x² - 5x) = 12x - 2x².

   A = ∫₀⁶ (12x - 2x²)dx = [6x² - 2x³/3]₀⁶ = 216 - 144

3. So. The enclosed area is 72 square units. You can check it with the shortcut |a|(β - α)³/6 = (2 × 6³)/6 = 72. The answer is 72.

The common slip: 36. Using the shortcut (β-α)³/6 without the leading coefficient |a| = 2 of 12x - 2x² gives 36, not 72.

The idea to keep. For two curves, integrate (upper minus lower) between the points where they meet. If that difference is a quadratic ax²+bx+c with roots α, β, the area is |a|(β-α)³/6.

Answer from the NTA final answer key. Solution by PrepWiser.