Application of Integrals: Diagram-Based Questions (7)

easy • 2 min read
Tags Application Of Integrals

Question

Find the area of the region bounded by the curve y=x2y = x^2 and the line y=x+2y = x + 2.

Solution — Step by Step

Set x2=x+2⇒x2−x−2=0⇒(x−2)(x+1)=0x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0, so x=−1,2x = -1, 2.

Between x=−1x = -1 and x=2x = 2, the line is above the parabola. Test x=0x = 0: line gives y=2y = 2, parabola gives y=0y = 0. Line wins.

 A=∫−12[(x+2)−x2] dxA = \int_{-1}^{2} [(x+2) - x^2]\, dx A=∫−12(x+2−x2) dx=[x22+2x−x33]−12A = \int_{-1}^{2} (x + 2 - x^2)\, dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}

At x=2x = 2: 2+4−8/3=6−8/3=10/32 + 4 - 8/3 = 6 - 8/3 = 10/3. At x=−1x = -1: 1/2−2+1/3=−7/61/2 - 2 + 1/3 = -7/6.

A=10/3−(−7/6)=20/6+7/6=27/6=9/2A = 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2

Final answer: A=9/2A = 9/2 square units.

Why This Works

The area between two curves is ∫ab[top−bottom] dx\int_a^b [\text{top} - \text{bottom}]\, dx where aa and bb are intersection points. Always sketch first to know which is on top — sign matters.

If the curves cross multiple times, split the integral at each crossing and switch top/bottom as needed.

Alternative Method

Integrate with respect to yy. Solve for xx from each curve: x=±yx = \pm\sqrt{y} (parabola) and x=y−2x = y - 2 (line). Limits and “right minus left” geometry — same answer but more complex when the parabola opens upward.

Forgetting to draw the sketch and assuming the parabola is on top. Always check at one interior point: at x=0x = 0, parabola =0= 0 and line =2= 2, so line is higher. Without this check, you might compute −9/2-9/2 and “correct” by absolute value — but on an exam, signed setups can cost a mark.

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