The question
Find the area of the region bounded by the curves x 2 = y x^2 = y x 2 = y y = x + 2 y = x + 2 y = x + 2 x x x
Exam answer Show all explanations
x 2 = y x^2 = y x 2 = y y = x + 2 y = x + 2 y = x + 2 x x x ( − 2 , 0 ) (-2, 0) ( − 2 , 0 )
Why?
y = x 2 y = x^2 y = x 2 x x x y y y
The line meets the x x x y = 0 y = 0 y = 0 x + 2 = 0 x + 2 = 0 x + 2 = 0 x = − 2 x = -2 x = − 2
Putting y = x + 2 y = x + 2 y = x + 2 x 2 = y x^2 = y x 2 = y
x 2 − x − 2 = 0 ⇒ ( x − 2 ) ( x + 1 ) = 0 ⇒ x = − 1 or x = 2 \begin{aligned} &x^2 - x - 2 = 0 \\ \Rightarrow\ &(x - 2)(x + 1) = 0 \\ \Rightarrow\ &x = -1 \text{ or } x = 2 \end{aligned} ⇒ ⇒ x 2 − x − 2 = 0 ( x − 2 ) ( x + 1 ) = 0 x = − 1 or x = 2 Why?
At a meeting point both equations hold, so x 2 = x + 2 x^2 = x + 2 x 2 = x + 2
To factorise, find two numbers with product − 2 -2 − 2 − 1 -1 − 1 − 2 -2 − 2 1 1 1
x 2 = x + 2 x 2 − x − 2 = 0 x 2 − 2 x + x − 2 = 0 x ( x − 2 ) + 1 ( x − 2 ) = 0 ( x − 2 ) ( x + 1 ) = 0 \begin{aligned} x^2 &= x + 2 \\ x^2 - x - 2 &= 0 \\ x^2 - 2x + x - 2 &= 0 \\ x(x - 2) + 1(x - 2) &= 0 \\ (x - 2)(x + 1) &= 0 \end{aligned} x 2 x 2 − x − 2 x 2 − 2 x + x − 2 x ( x − 2 ) + 1 ( x − 2 ) ( x − 2 ) ( x + 1 ) = x + 2 = 0 = 0 = 0 = 0 So x − 2 = 0 x - 2 = 0 x − 2 = 0 x = 2 x = 2 x = 2 x + 1 = 0 x + 1 = 0 x + 1 = 0 x = − 1 x = -1 x = − 1
So the x x x − 1 -1 − 1 2 2 2 ( − 1 , 1 ) (-1, 1) ( − 1 , 1 ) ( 2 , 4 ) (2, 4) ( 2 , 4 )
Why?
Find each y y y y = x + 2 y = x + 2 y = x + 2 − 1 + 2 = 1 -1 + 2 = 1 − 1 + 2 = 1 2 + 2 = 4 2 + 2 = 4 2 + 2 = 4 ( − 1 ) 2 = 1 (-1)^2 = 1 ( − 1 ) 2 = 1 2 2 = 4 2^2 = 4 2 2 = 4
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(−2, 0)
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x² = y
y = x + 2
The region bounded by y = x + 2 y = x + 2 y = x + 2 x = − 2 x = -2 x = − 2 x = − 1 x = -1 x = − 1 x 2 = y x^2 = y x 2 = y x = − 1 x = -1 x = − 1 x = 0 x = 0 x = 0 x x x The region lies above the x x x x = − 2 x = -2 x = − 2 x = − 1 x = -1 x = − 1 x = − 1 x = -1 x = − 1 x = 0 x = 0 x = 0
Why?
The region bounded by all three sits above the x x x
For − 2 ≤ x ≤ − 1 -2 \le x \le -1 − 2 ≤ x ≤ − 1 x + 2 ≤ x 2 x + 2 \le x^2 x + 2 ≤ x 2 − 1 ≤ x ≤ 0 -1 \le x \le 0 − 1 ≤ x ≤ 0 x = − 1 x = -1 x = − 1
Watch out: the region between the line and the parabola, from x = − 1 x = -1 x = − 1 x = 2 x = 2 x = 2 x x x
Required area = ∫ − 2 − 1 ( x + 2 ) d x + ∫ − 1 0 x 2 d x \begin{aligned} &\text{Required area} \\ &= \int_{-2}^{-1}(x + 2)\,dx + \int_{-1}^{0}x^2\,dx \end{aligned} Required area = ∫ − 2 − 1 ( x + 2 ) d x + ∫ − 1 0 x 2 d x Why?
The area under a curve y = f ( x ) ≥ 0 y = f(x) \ge 0 y = f ( x ) ≥ 0 x = a x = a x = a x = b x = b x = b
Area = ∫ a b y d x \text{Area} = \int_a^b y\,dx Area = ∫ a b y d x Use y = x + 2 y = x + 2 y = x + 2 y = x 2 y = x^2 y = x 2
= [ ( x + 2 ) 2 2 ] − 2 − 1 + [ x 3 3 ] − 1 0 \begin{aligned} &= \left[\frac{(x + 2)^2}{2}\right]_{-2}^{-1} + \left[\frac{x^3}{3}\right]_{-1}^{0} \end{aligned} = [ 2 ( x + 2 ) 2 ] − 2 − 1 + [ 3 x 3 ] − 1 0 Why?
Standard integrals:
∫ ( x + 2 ) d x = ( x + 2 ) 2 2 \int(x + 2)\,dx = \frac{(x + 2)^2}{2} ∫ ( x + 2 ) d x = 2 ( x + 2 ) 2 ∫ x 2 d x = x 3 3 \int x^2\,dx = \frac{x^3}{3} ∫ x 2 d x = 3 x 3 = ( 1 2 − 0 ) + ( 0 + 1 3 ) = 1 2 + 1 3 \begin{aligned} &= \left(\frac12 - 0\right) + \left(0 + \frac13\right) \\ &= \frac12 + \frac13 \end{aligned} = ( 2 1 − 0 ) + ( 0 + 3 1 ) = 2 1 + 3 1 Why?
In each bracket, upper limit minus lower limit:
( − 1 + 2 ) 2 2 − ( − 2 + 2 ) 2 2 = 1 2 0 3 3 − ( − 1 ) 3 3 = 1 3 \begin{aligned} \frac{(-1 + 2)^2}{2} - \frac{(-2 + 2)^2}{2} &= \frac12 \\ \frac{0^3}{3} - \frac{(-1)^3}{3} &= \frac13 \end{aligned} 2 ( − 1 + 2 ) 2 − 2 ( − 2 + 2 ) 2 3 0 3 − 3 ( − 1 ) 3 = 2 1 = 3 1 Step by step:
( − 1 + 2 ) 2 2 = 1 2 2 = 1 2 ( − 2 + 2 ) 2 2 = 0 2 2 = 0 0 3 3 = 0 ( − 1 ) 3 3 = − 1 3 0 − ( − 1 3 ) = 1 3 \begin{aligned} &\frac{(-1 + 2)^2}{2} = \frac{1^2}{2} = \frac12 \\ &\frac{(-2 + 2)^2}{2} = \frac{0^2}{2} = 0 \\[4pt] &\frac{0^3}{3} = 0 \\ &\frac{(-1)^3}{3} = \frac{-1}{3} \\ &0 - \left(-\frac13\right) = \frac13 \end{aligned} 2 ( − 1 + 2 ) 2 = 2 1 2 = 2 1 2 ( − 2 + 2 ) 2 = 2 0 2 = 0 3 0 3 = 0 3 ( − 1 ) 3 = 3 − 1 0 − ( − 3 1 ) = 3 1 Then 1 2 + 1 3 = 3 6 + 2 6 = 5 6 \frac12 + \frac13 = \frac36 + \frac26 = \frac56 2 1 + 3 1 = 6 3 + 6 2 = 6 5
Check: the first piece is a triangle with base 1 and height 1, so its area is 1 2 \frac12 2 1
∴ \therefore ∴ = 5 6 = \dfrac56 = 6 5
Where this comes from Chapter 8, Application of Integrals : Area bounded by a parabola, a line and the x-axis.
Question type: A curve and a line
How to spot it: click to see Type 7. A curve and a line: top minus bottom
Spot it: the region is shut in by a curve and a line, like a parabola and y = x y = x y = x ≤ \le ≤
Here: The parabola x 2 = y x^2 = y x 2 = y y = x + 2 y = x + 2 y = x + 2 x = − 1 x = -1 x = − 1 x x x x = − 1 x = -1 x = − 1 − 2 -2 − 2 − 1 -1 − 1 − 1 -1 − 1
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 7, Integrals .