The question
(a) Sketch the graph y = ∣ x + 1 ∣ y = |x + 1| y = ∣ x + 1∣ ∫ − 4 2 ∣ x + 1 ∣ d x \displaystyle\int_{-4}^{2}|x + 1|\,dx ∫ − 4 2 ∣ x + 1∣ d x
OR
(b) Using integration find the area of the region { ( x , y ) : x 2 − 4 y ≤ 0 , y − x ≤ 0 } \{(x, y) : x^2 - 4y \le 0, y - x \le 0\} {( x , y ) : x 2 − 4 y ≤ 0 , y − x ≤ 0 }
For Visually Impaired:
(a) Define the function y = ∣ x + 1 ∣ y = |x + 1| y = ∣ x + 1∣ ∫ − 4 2 ∣ x + 1 ∣ d x \displaystyle\int_{-4}^{2}|x + 1|\,dx ∫ − 4 2 ∣ x + 1∣ d x
OR
(b) Using integration find the area enclosed within the curve: 25 x 2 + 16 y 2 = 400 25x^2 + 16y^2 = 400 25 x 2 + 16 y 2 = 400
Exam answer Show all explanations
(a)
y = ∣ x + 1 ∣ = − x − 1 y = |x + 1| = -x - 1 y = ∣ x + 1∣ = − x − 1 x < − 1 x < -1 x < − 1 x + 1 x + 1 x + 1 x ≥ − 1 x \ge -1 x ≥ − 1 ( − 1 , 0 ) (-1, 0) ( − 1 , 0 )
Why?
By the definition of modulus, ∣ u ∣ = u |u| = u ∣ u ∣ = u u ≥ 0 u \ge 0 u ≥ 0 ∣ u ∣ = − u |u| = -u ∣ u ∣ = − u u < 0 u < 0 u < 0
Here u = x + 1 u = x + 1 u = x + 1 x < − 1 x < -1 x < − 1 − 1 -1 − 1 y = − x − 1 y = -x - 1 y = − x − 1 − 1 -1 − 1 y = x + 1 y = x + 1 y = x + 1 x = − 1 x = -1 x = − 1
x = −4
x = 2
y = |x + 1|
−4
−1
2
O
3
X
Y
The graph of y = ∣ x + 1 ∣ y = |x + 1| y = ∣ x + 1∣ x = − 4 x = -4 x = − 4 x = 2 x = 2 x = 2 ∫ − 4 2 ∣ x + 1 ∣ d x = ∫ − 4 − 1 ( − x − 1 ) d x + ∫ − 1 2 ( x + 1 ) d x \begin{aligned} &\int_{-4}^{2}|x + 1|\,dx \\ &= \int_{-4}^{-1}(-x - 1)\,dx \\ &\quad + \int_{-1}^{2}(x + 1)\,dx \end{aligned} ∫ − 4 2 ∣ x + 1∣ d x = ∫ − 4 − 1 ( − x − 1 ) d x + ∫ − 1 2 ( x + 1 ) d x Why?
A modulus cannot be integrated in one piece. Split the interval at x = − 1 x = -1 x = − 1 x + 1 x + 1 x + 1
∫ a b f d x = ∫ a c f d x + ∫ c b f d x \int_a^bf\,dx = \int_a^cf\,dx + \int_c^bf\,dx ∫ a b f d x = ∫ a c f d x + ∫ c b f d x ∫ − 4 2 ∣ x + 1 ∣ d x = − [ ( x + 1 ) 2 2 ] − 4 − 1 + [ ( x + 1 ) 2 2 ] − 1 2 = − ( 0 − 9 2 ) + ( 9 2 − 0 ) = 9 \begin{aligned} &\int_{-4}^{2}|x + 1|\,dx \\ &= -\left[\frac{(x + 1)^2}{2}\right]_{-4}^{-1} \\ &\quad + \left[\frac{(x + 1)^2}{2}\right]_{-1}^{2} \\ &= -\left(0 - \frac92\right) + \left(\frac92 - 0\right) = 9 \end{aligned} ∫ − 4 2 ∣ x + 1∣ d x = − [ 2 ( x + 1 ) 2 ] − 4 − 1 + [ 2 ( x + 1 ) 2 ] − 1 2 = − ( 0 − 2 9 ) + ( 2 9 − 0 ) = 9 Why?
Since d d x ( x + 1 ) = 1 \frac{d}{dx}(x + 1) = 1 d x d ( x + 1 ) = 1 x + 1 x + 1 x + 1 ( x + 1 ) 2 2 \frac{(x + 1)^2}{2} 2 ( x + 1 ) 2 − x − 1 = − ( x + 1 ) -x - 1 = -(x + 1) − x − 1 = − ( x + 1 )
Values of ( x + 1 ) 2 2 \frac{(x + 1)^2}{2} 2 ( x + 1 ) 2
x = − 1 : 0 2 2 = 0 x = − 4 : ( − 3 ) 2 2 = 9 2 x = 2 : 3 2 2 = 9 2 \begin{aligned} x = -1&: \frac{0^2}{2} = 0 \\ x = -4&: \frac{(-3)^2}{2} = \frac92 \\ x = 2&: \frac{3^2}{2} = \frac92 \end{aligned} x = − 1 x = − 4 x = 2 : 2 0 2 = 0 : 2 ( − 3 ) 2 = 2 9 : 2 3 2 = 2 9 (At x = − 4 x = -4 x = − 4 x + 1 = − 4 + 1 = − 3 x + 1 = -4 + 1 = -3 x + 1 = − 4 + 1 = − 3 x = 2 x = 2 x = 2 x + 1 = 2 + 1 = 3 x + 1 = 2 + 1 = 3 x + 1 = 2 + 1 = 3
Each bracket is (value at the top limit) minus (value at the bottom limit):
− ( 0 − 9 2 ) + ( 9 2 − 0 ) = − ( − 9 2 ) + 9 2 = 9 2 + 9 2 = 18 2 = 9 \begin{aligned} &-\left(0 - \frac92\right) + \left(\frac92 - 0\right) \\ &= -\left(-\frac92\right) + \frac92 \\ &= \frac92 + \frac92 = \frac{18}{2} = 9 \end{aligned} − ( 0 − 2 9 ) + ( 2 9 − 0 ) = − ( − 2 9 ) + 2 9 = 2 9 + 2 9 = 2 18 = 9 ∫ − 4 2 ∣ x + 1 ∣ d x = 9 \int_{-4}^{2}|x + 1|\,dx = 9 ∫ − 4 2 ∣ x + 1∣ d x = 9 y = ∣ x + 1 ∣ y = |x + 1| y = ∣ x + 1∣ x = − 4 x = -4 x = − 4 x = 2 x = 2 x = 2
Why?
∣ x + 1 ∣ ≥ 0 |x + 1| \ge 0 ∣ x + 1∣ ≥ 0
Check with the figure: the region is two triangles, each with base 3 and height 3, so each has area 1 2 × 3 × 3 = 9 2 \frac12 \times 3 \times 3 = \frac92 2 1 × 3 × 3 = 2 9
(b)
The region lies on or above the parabola x 2 = 4 y x^2 = 4y x 2 = 4 y y = x y = x y = x
Why?
Read each inequality as a side of a curve:
x 2 − 4 y ≤ 0 ⟺ y ≥ x 2 4 y − x ≤ 0 ⟺ y ≤ x \begin{aligned} x^2 - 4y \le 0 &\iff y \ge \frac{x^2}{4} \\ y - x \le 0 &\iff y \le x \end{aligned} x 2 − 4 y ≤ 0 y − x ≤ 0 ⟺ y ≥ 4 x 2 ⟺ y ≤ x So the region is squeezed between the parabola (below) and the line (above).
Points of intersection: x 2 = 4 x ⇒ x = 0 x^2 = 4x \Rightarrow x = 0 x 2 = 4 x ⇒ x = 0 x = 4 x = 4 x = 4 ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 4 , 4 ) (4, 4) ( 4 , 4 )
Why?
At a meeting point both equations hold, so put y = x y = x y = x x 2 = 4 y x^2 = 4y x 2 = 4 y
x 2 − 4 x = x ( x − 4 ) = 0 x^2 - 4x = x(x - 4) = 0 x 2 − 4 x = x ( x − 4 ) = 0 Watch out: do not divide by x x x x = 0 x = 0 x = 0 y = x y = x y = x ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 4 , 4 ) (4, 4) ( 4 , 4 )
O
(4, 4)
x² = 4y
y = x
4
4
X
Y
The region between the line y = x y = x y = x x 2 = 4 y x^2 = 4y x 2 = 4 y x = 0 x = 0 x = 0 x = 4 x = 4 x = 4 Area = ∫ 0 4 x d x − ∫ 0 4 x 2 4 d x \begin{aligned} \text{Area} &= \int_0^4x\,dx - \int_0^4\frac{x^2}{4}\,dx \end{aligned} Area = ∫ 0 4 x d x − ∫ 0 4 4 x 2 d x Why?
The area between two curves is the integral of (upper curve minus lower curve) between the meeting points:
Area = ∫ a b ( y upper − y lower ) d x \text{Area} = \int_a^b(y_{\text{upper}} - y_{\text{lower}})\,dx Area = ∫ a b ( y upper − y lower ) d x On [ 0 , 4 ] [0, 4] [ 0 , 4 ] x = 2 x = 2 x = 2 4 4 = 1 \frac{4}{4} = 1 4 4 = 1 x 2 = 4 y x^2 = 4y x 2 = 4 y y = x 2 4 y = \frac{x^2}{4} y = 4 x 2
Area = [ x 2 2 ] 0 4 − 1 12 [ x 3 ] 0 4 = 1 2 ( 16 − 0 ) − 1 12 ( 64 − 0 ) = 8 − 16 3 \begin{aligned} \text{Area} &= \left[\frac{x^2}{2}\right]_0^4 - \frac{1}{12}\Big[x^3\Big]_0^4 \\ &= \frac12(16 - 0) \\ &\quad - \frac{1}{12}(64 - 0) \\ &= 8 - \frac{16}{3} \end{aligned} Area = [ 2 x 2 ] 0 4 − 12 1 [ x 3 ] 0 4 = 2 1 ( 16 − 0 ) − 12 1 ( 64 − 0 ) = 8 − 3 16 Why?
Power rule: ∫ x d x = x 2 2 \int x\,dx = \frac{x^2}{2} ∫ x d x = 2 x 2 ∫ x 2 4 d x = 1 4 ⋅ x 3 3 = x 3 12 \int\frac{x^2}{4}\,dx = \frac14\cdot\frac{x^3}{3} = \frac{x^3}{12} ∫ 4 x 2 d x = 4 1 ⋅ 3 x 3 = 12 x 3
At x = 4 x = 4 x = 4 4 2 = 16 4^2 = 16 4 2 = 16 4 3 = 64 4^3 = 64 4 3 = 64 64 12 = 16 3 \frac{64}{12} = \frac{16}{3} 12 64 = 3 16
1 2 ( 16 ) = 16 2 = 8 1 12 ( 64 ) = 64 12 = 64 ÷ 4 12 ÷ 4 = 16 3 8 = 8 × 3 3 = 24 3 \begin{aligned} \frac12(16) &= \frac{16}{2} = 8 \\ \frac{1}{12}(64) &= \frac{64}{12} = \frac{64 \div 4}{12 \div 4} = \frac{16}{3} \\[4pt] 8 &= \frac{8 \times 3}{3} = \frac{24}{3} \end{aligned} 2 1 ( 16 ) 12 1 ( 64 ) 8 = 2 16 = 8 = 12 64 = 12 ÷ 4 64 ÷ 4 = 3 16 = 3 8 × 3 = 3 24 8 − 16 3 = 24 − 16 3 = 8 3 8 - \frac{16}{3} = \frac{24 - 16}{3} = \frac83 8 − 3 16 = 3 24 − 16 = 3 8 ∴ \therefore ∴ = 8 3 = \frac83 = 3 8
For Visually Impaired
(a)
y = ∣ x + 1 ∣ = { − x − 1 , x < − 1 x + 1 , x ≥ − 1 y = |x + 1| = \begin{cases} -x - 1, & x < -1 \\ x + 1, & x \ge -1 \end{cases} y = ∣ x + 1∣ = { − x − 1 , x + 1 , x < − 1 x ≥ − 1 Why?
By the definition of modulus, ∣ u ∣ = u |u| = u ∣ u ∣ = u u ≥ 0 u \ge 0 u ≥ 0 ∣ u ∣ = − u |u| = -u ∣ u ∣ = − u u < 0 u < 0 u < 0 u = x + 1 u = x + 1 u = x + 1 x < − 1 x < -1 x < − 1
∫ − 4 2 ∣ x + 1 ∣ d x = ∫ − 4 − 1 ( − x − 1 ) d x + ∫ − 1 2 ( x + 1 ) d x \begin{aligned} &\int_{-4}^{2}|x + 1|\,dx \\ &= \int_{-4}^{-1}(-x - 1)\,dx \\ &\quad + \int_{-1}^{2}(x + 1)\,dx \end{aligned} ∫ − 4 2 ∣ x + 1∣ d x = ∫ − 4 − 1 ( − x − 1 ) d x + ∫ − 1 2 ( x + 1 ) d x Why?
Split the interval at x = − 1 x = -1 x = − 1 x + 1 x + 1 x + 1
∫ − 4 2 ∣ x + 1 ∣ d x = − [ ( x + 1 ) 2 2 ] − 4 − 1 + [ ( x + 1 ) 2 2 ] − 1 2 = − ( 0 − 9 2 ) + ( 9 2 − 0 ) = 9 \begin{aligned} &\int_{-4}^{2}|x + 1|\,dx \\ &= -\left[\frac{(x + 1)^2}{2}\right]_{-4}^{-1} \\ &\quad + \left[\frac{(x + 1)^2}{2}\right]_{-1}^{2} \\ &= -\left(0 - \frac92\right) + \left(\frac92 - 0\right) = 9 \end{aligned} ∫ − 4 2 ∣ x + 1∣ d x = − [ 2 ( x + 1 ) 2 ] − 4 − 1 + [ 2 ( x + 1 ) 2 ] − 1 2 = − ( 0 − 2 9 ) + ( 2 9 − 0 ) = 9 Why?
An antiderivative of x + 1 x + 1 x + 1 ( x + 1 ) 2 2 \frac{(x + 1)^2}{2} 2 ( x + 1 ) 2 − ( x + 1 ) -(x + 1) − ( x + 1 )
At x = − 1 x = -1 x = − 1 x = − 4 x = -4 x = − 4 ( − 3 ) 2 2 = 9 2 \frac{(-3)^2}{2} = \frac92 2 ( − 3 ) 2 = 2 9 x = 2 x = 2 x = 2 3 2 2 = 9 2 \frac{3^2}{2} = \frac92 2 3 2 = 2 9
− ( 0 − 9 2 ) + ( 9 2 − 0 ) = − ( − 9 2 ) + 9 2 = 9 2 + 9 2 = 18 2 = 9 \begin{aligned} &-\left(0 - \frac92\right) + \left(\frac92 - 0\right) \\ &= -\left(-\frac92\right) + \frac92 \\ &= \frac92 + \frac92 = \frac{18}{2} = 9 \end{aligned} − ( 0 − 2 9 ) + ( 2 9 − 0 ) = − ( − 2 9 ) + 2 9 = 2 9 + 2 9 = 2 18 = 9 It represents the area of the region bounded by the curve y = ∣ x + 1 ∣ y = |x + 1| y = ∣ x + 1∣ x = − 4 x = -4 x = − 4 x = 2 x = 2 x = 2
Why?
∣ x + 1 ∣ |x + 1| ∣ x + 1∣ 1 2 × 3 × 3 = 9 2 \frac12 \times 3 \times 3 = \frac92 2 1 × 3 × 3 = 2 9
(b)
25 x 2 + 16 y 2 = 400 ⇒ x 2 16 + y 2 25 = 1 ⇒ y = 5 4 4 2 − x 2 \begin{aligned} &25x^2 + 16y^2 = 400 \\ \Rightarrow\ &\frac{x^2}{16} + \frac{y^2}{25} = 1 \\ \Rightarrow\ &y = \frac54\sqrt{4^2 - x^2} \end{aligned} ⇒ ⇒ 25 x 2 + 16 y 2 = 400 16 x 2 + 25 y 2 = 1 y = 4 5 4 2 − x 2 Why?
Divide by 400 to get the standard form x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 a = 4 a = 4 a = 4 b = 5 b = 5 b = 5
25 x 2 400 + 16 y 2 400 = 400 400 x 2 16 + y 2 25 = 1 \begin{aligned} \frac{25x^2}{400} + \frac{16y^2}{400} &= \frac{400}{400} \\ \frac{x^2}{16} + \frac{y^2}{25} &= 1 \end{aligned} 400 25 x 2 + 400 16 y 2 16 x 2 + 25 y 2 = 400 400 = 1 (400 ÷ 25 = 16 400 \div 25 = 16 400 ÷ 25 = 16 400 ÷ 16 = 25 400 \div 16 = 25 400 ÷ 16 = 25
Then solve for y y y
y 2 25 = 1 − x 2 16 y 2 = 25 ( 1 − x 2 16 ) = 25 ⋅ 16 − x 2 16 = 25 16 ( 16 − x 2 ) \begin{aligned} \frac{y^2}{25} &= 1 - \frac{x^2}{16} \\ y^2 &= 25\left(1 - \frac{x^2}{16}\right) \\ &= 25 \cdot \frac{16 - x^2}{16} \\ &= \frac{25}{16}(16 - x^2) \end{aligned} 25 y 2 y 2 = 1 − 16 x 2 = 25 ( 1 − 16 x 2 ) = 25 ⋅ 16 16 − x 2 = 16 25 ( 16 − x 2 ) The square root of 25 16 \frac{25}{16} 16 25 5 4 \frac54 4 5
y = 25 16 16 − x 2 = 5 4 4 2 − x 2 \begin{aligned} y &= \sqrt{\frac{25}{16}}\sqrt{16 - x^2} \\ &= \frac54\sqrt{4^2 - x^2} \end{aligned} y = 16 25 16 − x 2 = 4 5 4 2 − x 2 This is the part in the first quadrant (y ≥ 0 y \ge 0 y ≥ 0
Why?
The four quarters of the ellipse have equal areas, so the whole area is 4 times the area in the first quadrant.
In the first quadrant x x x a = 4 a = 4 a = 4 y ≥ 0 y \ge 0 y ≥ 0
Area = 4 ∫ 0 4 5 4 4 2 − x 2 d x = 5 [ x 4 2 − x 2 2 + 4 2 2 sin − 1 x 4 ] 0 4 \begin{aligned} \text{Area} &= 4\int_0^4\frac54\sqrt{4^2 - x^2}\,dx \\ &= 5\bigg[\frac{x\sqrt{4^2 - x^2}}{2} \\ &\qquad + \frac{4^2}{2}\sin^{-1}\frac{x}{4}\bigg]_0^4 \end{aligned} Area = 4 ∫ 0 4 4 5 4 2 − x 2 d x = 5 [ 2 x 4 2 − x 2 + 2 4 2 sin − 1 4 x ] 0 4 Why?
The constants come out first: 4 × 5 4 = 5 4 \times \frac54 = 5 4 × 4 5 = 5
∫ a 2 − x 2 d x = x a 2 − x 2 2 + a 2 2 sin − 1 x a + C \begin{aligned} &\int\sqrt{a^2 - x^2}\,dx \\ &= \frac{x\sqrt{a^2 - x^2}}{2} \\ &\quad + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \end{aligned} ∫ a 2 − x 2 d x = 2 x a 2 − x 2 + 2 a 2 sin − 1 a x + C with a = 4 a = 4 a = 4
Area = 5 [ 0 + 8 sin − 1 ( 1 ) − 0 ] = 40 × π 2 \begin{aligned} \text{Area} &= 5[0 + 8\sin^{-1}(1) - 0] \\ &= 40 \times \frac{\pi}{2} \end{aligned} Area = 5 [ 0 + 8 sin − 1 ( 1 ) − 0 ] = 40 × 2 π Why?
At x = 4 x = 4 x = 4 16 − 16 = 0 \sqrt{16 - 16} = 0 16 − 16 = 0 4 2 2 = 8 \frac{4^2}{2} = 8 2 4 2 = 8
At x = 0 x = 0 x = 0 sin − 1 0 = 0 \sin^{-1}0 = 0 sin − 1 0 = 0
Then sin − 1 1 = π 2 \sin^{-1}1 = \frac{\pi}{2} sin − 1 1 = 2 π 40 × π 2 = 20 π 40 \times \frac{\pi}{2} = 20\pi 40 × 2 π = 20 π
x = 4 : 4 16 − 16 2 + 8 sin − 1 4 4 = 4 × 0 2 + 8 sin − 1 1 = 0 + 8 sin − 1 1 x = 0 : 0 ⋅ 16 2 + 8 sin − 1 0 = 0 + 0 = 0 5 × 8 sin − 1 1 = 40 × π 2 = 40 π 2 = 20 π \begin{aligned} x = 4&: \frac{4\sqrt{16 - 16}}{2} + 8\sin^{-1}\frac44 \\ &= \frac{4 \times 0}{2} + 8\sin^{-1}1 \\ &= 0 + 8\sin^{-1}1 \\[4pt] x = 0&: \frac{0 \cdot \sqrt{16}}{2} + 8\sin^{-1}0 \\ &= 0 + 0 = 0 \\[4pt] 5 \times 8\sin^{-1}1 &= 40 \times \frac{\pi}{2} \\ &= \frac{40\pi}{2} = 20\pi \end{aligned} x = 4 x = 0 5 × 8 sin − 1 1 : 2 4 16 − 16 + 8 sin − 1 4 4 = 2 4 × 0 + 8 sin − 1 1 = 0 + 8 sin − 1 1 : 2 0 ⋅ 16 + 8 sin − 1 0 = 0 + 0 = 0 = 40 × 2 π = 2 40 π = 20 π ∴ \therefore ∴ = 20 π = 20\pi = 20 π
Where this comes from Chapter 8, Application of Integrals : Area by integration: |x + 1|, parabola and line.
Question type: (a) A modulus inside the integral ; (b) A curve and a line
How to spot it: click to see (a) Type 6. A modulus inside the integral: split at the corner
Spot it: bars inside the integral, like ∣ x + 3 ∣ |x + 3| ∣ x + 3∣ x ∣ x ∣ x|x| x ∣ x ∣
Here: ∣ x + 1 ∣ |x + 1| ∣ x + 1∣ x = − 1 x = -1 x = − 1 − x − 1 -x - 1 − x − 1 − 4 -4 − 4 − 1 -1 − 1 x + 1 x + 1 x + 1 − 1 -1 − 1 x x x
(b) 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: above x 2 = 4 y x^2 = 4y x 2 = 4 y y = x y = x y = x ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 4 , 4 ) (4, 4) ( 4 , 4 ) ∫ 0 4 ( x − x 2 4 ) d x \int_0^4\left(x - \frac{x^2}{4}\right)dx ∫ 0 4 ( x − 4 x 2 ) d x
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 7, Integrals .