f(x)=x|x|-1 { and } g(x)={cases} 3x/2, & x>0 \\ 2x, & x≤ 0 {cases} What is the area bounded by the curves?

Correct Answer is : 838/338 square units

Correct Answer is : 838/338 square units

Test Index

UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 12-13

For the next two (2) items that follow:

Consider the curves

Question : 13 of 120

Marks: +1, -0

f(x)=x|x|-1 \text{ and } g(x)=\begin{cases} \frac{3x}{2}, & x>0 \\ 2x, & x\le 0 \end{cases}

What is the area bounded by the curves?

Solution:

Concept:

Area between two curves is

\int_{a}^{b}[\text{top curve} - \text{bottom curve}]\,dx

Explanation:

Given

f(x)=x|x|-1

and

g(x)=\begin{cases} \frac{3x}{2}, & x>0 \\ 2x, & x\le 0 \end{cases}

Find intersection points:

(-1,-2)

and

(2,3)

Note that

f(x)

changes at

x=0

due to

|x|

: for

x\ge 0

f(x)=x^2-1

; for

x<0

f(x)=-x^2-1

Thus split the area into two intervals:

[-1,0]

and

[0,2]

[-1,0]

: top is

g(x)=2x

, bottom is

f(x)=-x^2-1

Area part =

\int_{-1}^0 [2x - (-x^2-1)]\,dx = \int_{-1}^0 (x^2+2x+1)\,dx

Compute:

\left[\frac{x^3}{3}+x^2+x\right]_{-1}^0 = (0) - \left(-\frac13+1-1\right) = \frac13

[0,2]

: top is

g(x)=\frac{3x}{2}

, bottom is

f(x)=x^2-1

Area part =

\int_0^2 \left[\frac{3x}{2} - (x^2-1)\right] dx = \int_0^2 \left(\frac{3x}{2} - x^2 +1\right) dx

Compute:

\left[\frac{3x^2}{4} - \frac{x^3}{3} + x\right]_0^2 = \left(3 - \frac{8}{3} + 2\right) - 0 = \frac{7}{3}

Total area =

\frac13 + \frac{7}{3} = \frac{8}{3}

square units.

Answer:

\frac{8}{3}

square units.

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