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

Correct answer: B. (8)/(3) square units aligned Area & =∫_ -1 ^2[g(x)-f(x)] d x & =∫_ -1 ^0[g(x)-f(x)] d x+∫_0^2[g(x)-f(x)] d x & =∫_ -1 ^0(2 x+x^2+1) d x+∫_0^2[(3 x)/(2)-x^2+1] d x & =I_1+I_2 I_1 & =((2 x^2)/(2)+(1)/(3) x^3+x)_ -1 ^0 aligned aligned & =(0+0+0)-((-1)^2+(1)/(3)(-1)^3+(-1)) & =-(1-(1)/(3)-1)=(1)/(3) I_2 & =∫_0^2[(3 x)/(2)-x^2+1] d x & =[(3)/(2) (x^2)/(2)-(1)/(3) x^3+x]_0^2 & =((3)/(4) × 4-(1)/(3) × 8+2)-(0) & =5-(8)/(3)=(7)/(3) Area & =I_1+I_2=(1)/(3)+(7)/(3) aligned Area = (8)/(3) square units.

NDA 2016

Exam: NDA
Year: 2016
Subject: Mathematics
Chapter: Area Under Curves

NDA 2016 · Mathematics · Area Under Curves

Consider the curves f(x)=x|x|-1 and g(x)= cases (3 x)/(2), & x > 0...

Consider the curves

f(x)=x|x|-1

and

g(x)=

\begin{cases}\frac{3 x}{2}, & x > 0 \\ 2 x, & x \leq 0\end{cases}$$ What is the area bounded by the curves?

Options

A $\frac{17}{6}$ square units
B $\frac{8}{3}$ square units
C2 square units
D $\frac{1}{3}$ square unit

Correct answer

B. $\frac{8}{3}$ square units

Step-by-step solution

\begin{aligned} \text{ Area } & =\int_{-1}^2[g(x)-f(x)] d x \\ & =\int_{-1}^0[g(x)-f(x)] d x+\int_0^2[g(x)-f(x)] d x \\ & =\int_{-1}^0\left(2 x+x^2+1\right) d x+\int_0^2\left[\frac{3 x}{2}-x^2+1\right] d x \\ & =I_1+I_2 \\ I_1 & =\left(\frac{2 x^2}{2}+\frac{1}{3} x^3+x\right)_{-1}^0\end{aligned}

\begin{aligned} & =(0+0+0)-\left((-1)^2+\frac{1}{3}(-1)^3+(-1)\right) \\ & =-\left(1-\frac{1}{3}-1\right)=\frac{1}{3} \\ I_2 & =\int_0^2\left[\frac{3 x}{2}-x^2+1\right] d x \\ & =\left[\frac{3}{2} \frac{x^2}{2}-\frac{1}{3} x^3+x\right]_0^2 \\ & =\left(\frac{3}{4} \times 4-\frac{1}{3} \times 8+2\right)-(0) \\ & =5-\frac{8}{3}=\frac{7}{3} \\ \text{ Area } & =I_1+I_2=\frac{1}{3}+\frac{7}{3}\end{aligned}

Area =

\frac{8}{3} square units.$

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