The question
Make a rough sketch of the region { ( x , y ) : 0 ≤ y ≤ x 2 + 1 , 0 ≤ y ≤ x + 1 , 0 ≤ x ≤ 2 } \{(x, y) : 0 \le y \le x^2 + 1, 0 \le y \le x + 1, 0 \le x \le 2\} {( x , y ) : 0 ≤ y ≤ x 2 + 1 , 0 ≤ y ≤ x + 1 , 0 ≤ x ≤ 2 }
Exam answer Show all explanations
To find the points of intersection of the curve y = x 2 + 1 y = x^2 + 1 y = x 2 + 1 y = x + 1 y = x + 1 y = x + 1
x 2 + 1 = x + 1 ⇒ x ( x − 1 ) = 0 x^2 + 1 = x + 1 \Rightarrow x(x - 1) = 0 x 2 + 1 = x + 1 ⇒ x ( x − 1 ) = 0 Why?
Where the curve and the line meet, they have the same y y y
Subtract x + 1 x + 1 x + 1 x 2 − x = 0 x^2 - x = 0 x 2 − x = 0 x x x
x 2 + 1 = x + 1 x 2 + 1 − x − 1 = 0 x 2 − x = 0 x ( x − 1 ) = 0 \begin{aligned} x^2 + 1 &= x + 1 \\ x^2 + 1 - x - 1 &= 0 \\ x^2 - x &= 0 \\ x(x - 1) &= 0 \end{aligned} x 2 + 1 x 2 + 1 − x − 1 x 2 − x x ( x − 1 ) = x + 1 = 0 = 0 = 0 A product is 0 when one factor is 0: x = 0 x = 0 x = 0 x − 1 = 0 x - 1 = 0 x − 1 = 0 x = 1 x = 1 x = 1
Watch out: do not divide by x x x x = 0 x = 0 x = 0
⇒ x = 0 , 1 \Rightarrow x = 0, 1 ⇒ x = 0 , 1 So, the points of intersection are P ( 0 , 1 ) P(0, 1) P ( 0 , 1 ) Q ( 1 , 2 ) Q(1, 2) Q ( 1 , 2 )
Why?
Find each y y y y = x + 1 y = x + 1 y = x + 1 x = 0 x = 0 x = 0 y = 1 y = 1 y = 1 x = 1 x = 1 x = 1 y = 2 y = 2 y = 2
For 0 ≤ x ≤ 1 0 \le x \le 1 0 ≤ x ≤ 1 x 2 + 1 ≤ x + 1 x^2 + 1 \le x + 1 x 2 + 1 ≤ x + 1 1 ≤ x ≤ 2 1 \le x \le 2 1 ≤ x ≤ 2 x + 1 ≤ x 2 + 1 x + 1 \le x^2 + 1 x + 1 ≤ x 2 + 1 R ( 2 , 3 ) R(2, 3) R ( 2 , 3 ) S ( 1 , 0 ) S(1, 0) S ( 1 , 0 ) T ( 2 , 0 ) T(2, 0) T ( 2 , 0 )
Why?
A point of the region must have y y y x x x
Test a point in each part. At x = 1 2 x = \frac12 x = 2 1 x 2 + 1 = 5 4 x^2 + 1 = \frac54 x 2 + 1 = 4 5 x + 1 = 3 2 x + 1 = \frac32 x + 1 = 2 3 x = 3 2 x = \frac32 x = 2 3 x 2 + 1 = 13 4 x^2 + 1 = \frac{13}{4} x 2 + 1 = 4 13 x + 1 = 5 2 x + 1 = \frac52 x + 1 = 2 5
( 1 2 ) 2 + 1 = 1 4 + 4 4 = 5 4 1 2 + 1 = 1 2 + 2 2 = 3 2 = 6 4 ( 3 2 ) 2 + 1 = 9 4 + 4 4 = 13 4 3 2 + 1 = 3 2 + 2 2 = 5 2 = 10 4 \begin{aligned} \left(\tfrac12\right)^2 + 1 &= \tfrac14 + \tfrac44 = \tfrac54 \\ \tfrac12 + 1 &= \tfrac12 + \tfrac22 = \tfrac32 = \tfrac64 \\[4pt] \left(\tfrac32\right)^2 + 1 &= \tfrac94 + \tfrac44 = \tfrac{13}{4} \\ \tfrac32 + 1 &= \tfrac32 + \tfrac22 = \tfrac52 = \tfrac{10}{4} \end{aligned} ( 2 1 ) 2 + 1 2 1 + 1 ( 2 3 ) 2 + 1 2 3 + 1 = 4 1 + 4 4 = 4 5 = 2 1 + 2 2 = 2 3 = 4 6 = 4 9 + 4 4 = 4 13 = 2 3 + 2 2 = 2 5 = 4 10 Compare quarters: 5 4 < 6 4 \frac54 < \frac64 4 5 < 4 6 13 4 > 10 4 \frac{13}{4} > \frac{10}{4} 4 13 > 4 10
The bottom edge is the x x x y ≥ 0 y \ge 0 y ≥ 0 x = 0 x = 0 x = 0 x = 2 x = 2 x = 2 x = 2 x = 2 x = 2 R ( 2 , 3 ) R(2, 3) R ( 2 , 3 )
X
Y
O
P(0, 1)
Q(1, 2)
R(2, 3)
S(1, 0)
T(2, 0)
y = x² + 1
y = x + 1
Area of the shaded region OPQRTSO = area of OSQPO + area of STRQS.
Why?
The top edge changes from the parabola to the line at x = 1 x = 1 x = 1
The area between a graph y = f ( x ) ≥ 0 y = f(x) \ge 0 y = f ( x ) ≥ 0 x x x x = a x = a x = a x = b x = b x = b
Area = ∫ a b f ( x ) d x \text{Area} = \int_a^b f(x)\,dx Area = ∫ a b f ( x ) d x Watch out: this region is between the graphs and the x x x
Required area = ∫ 0 1 ( x 2 + 1 ) d x + ∫ 1 2 ( x + 1 ) d x = [ x 3 3 + x ] 0 1 + [ x 2 2 + x ] 1 2 \begin{aligned} &\text{Required area} \\ &= \int_0^1 (x^2 + 1)\,dx + \int_1^2 (x + 1)\,dx \\ &= \left[\frac{x^3}{3} + x\right]_0^1 + \left[\frac{x^2}{2} + x\right]_1^2 \end{aligned} Required area = ∫ 0 1 ( x 2 + 1 ) d x + ∫ 1 2 ( x + 1 ) d x = [ 3 x 3 + x ] 0 1 + [ 2 x 2 + x ] 1 2 Why?
Each term uses the power rule:
∫ x n d x = x n + 1 n + 1 \int x^n\,dx = \frac{x^{n + 1}}{n + 1} ∫ x n d x = n + 1 x n + 1 So x 2 x^2 x 2 x 3 3 \frac{x^3}{3} 3 x 3 x x x x 2 2 \frac{x^2}{2} 2 x 2 x x x
= [ ( 1 3 + 1 ) − 0 ] + [ ( 2 + 2 ) − ( 1 2 + 1 ) ] = 4 3 + 5 2 = 23 6 \begin{aligned} &= \left[\left(\frac13 + 1\right) - 0\right] \\ &\quad + \left[(2 + 2) - \left(\frac12 + 1\right)\right] \\ &= \frac43 + \frac52 = \frac{23}{6} \end{aligned} = [ ( 3 1 + 1 ) − 0 ] + [ ( 2 + 2 ) − ( 2 1 + 1 ) ] = 3 4 + 2 5 = 6 23 Why?
Each bracket is the value at the upper limit minus the value at the lower limit.
First: 1 3 + 1 = 4 3 \frac13 + 1 = \frac43 3 1 + 1 = 3 4
Second: at 2, 4 2 + 2 = 4 \frac{4}{2} + 2 = 4 2 4 + 2 = 4 1 2 + 1 = 3 2 \frac12 + 1 = \frac32 2 1 + 1 = 2 3 4 − 3 2 = 5 2 4 - \frac32 = \frac52 4 − 2 3 = 2 5
[ x 3 3 + x ] 0 1 = ( 1 3 3 + 1 ) − ( 0 3 3 + 0 ) = 1 3 + 3 3 − 0 = 4 3 [ x 2 2 + x ] 1 2 = ( 2 2 2 + 2 ) − ( 1 2 2 + 1 ) = ( 2 + 2 ) − ( 1 2 + 2 2 ) = 4 − 3 2 = 8 2 − 3 2 = 5 2 \begin{aligned} &\left[\frac{x^3}{3} + x\right]_0^1 \\ &= \left(\frac{1^3}{3} + 1\right) - \left(\frac{0^3}{3} + 0\right) \\ &= \frac13 + \frac33 - 0 = \frac43 \\[4pt] &\left[\frac{x^2}{2} + x\right]_1^2 \\ &= \left(\frac{2^2}{2} + 2\right) - \left(\frac{1^2}{2} + 1\right) \\ &= (2 + 2) - \left(\frac12 + \frac22\right) \\ &= 4 - \frac32 = \frac82 - \frac32 = \frac52 \end{aligned} [ 3 x 3 + x ] 0 1 = ( 3 1 3 + 1 ) − ( 3 0 3 + 0 ) = 3 1 + 3 3 − 0 = 3 4 [ 2 x 2 + x ] 1 2 = ( 2 2 2 + 2 ) − ( 2 1 2 + 1 ) = ( 2 + 2 ) − ( 2 1 + 2 2 ) = 4 − 2 3 = 2 8 − 2 3 = 2 5 Adding over the denominator 6: 8 6 + 15 6 = 23 6 \frac86 + \frac{15}{6} = \frac{23}{6} 6 8 + 6 15 = 6 23
4 3 = 4 × 2 3 × 2 = 8 6 5 2 = 5 × 3 2 × 3 = 15 6 8 6 + 15 6 = 8 + 15 6 = 23 6 \begin{aligned} \frac43 &= \frac{4 \times 2}{3 \times 2} = \frac86 \\ \frac52 &= \frac{5 \times 3}{2 \times 3} = \frac{15}{6} \\ \frac86 + \frac{15}{6} &= \frac{8 + 15}{6} = \frac{23}{6} \end{aligned} 3 4 2 5 6 8 + 6 15 = 3 × 2 4 × 2 = 6 8 = 2 × 3 5 × 3 = 6 15 = 6 8 + 15 = 6 23 Hence, the required area is 23 6 \dfrac{23}{6} 6 23
Where this comes from Chapter 8, Application of Integrals : Area of a region bounded by a parabola and a line.
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: Read the set: under both y = x 2 + 1 y = x^2 + 1 y = x 2 + 1 y = x + 1 y = x + 1 y = x + 1 x = 0 x = 0 x = 0 4 3 + 5 2 \frac43 + \frac52 3 4 + 2 5
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 7, Integrals .