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Area Under the Curves question

2023 · 8 Apr · Shift 2 · Q36
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  5. /2023 · 8 Apr · Shift 2 · Q36

Area Under the Curves question

2023 · 8 Apr · Shift 2 · Q36

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
Let the area enclosed by the lines x+y=2,y=0,x=0x+y=2, \mathrm{y}=0, x=0x+y=2,y=0,x=0 and the curve f(x)=min⁡{x2+34,1+[x]}f(x)=\min \left\{x^{2}+\frac{3}{4}, 1+[x]\right\}f(x)=min{x2+43​,1+[x]} where [x][x][x] denotes the greatest integer ≤x\leq x≤x, be A\mathrm{A}A. Then the value of 12 A12 \mathrm{~A}12 A is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 7

  1. Region enclosed by the given lines

The lines x+y=2, y=0, x=0x+y=2,\, y=0,\, x=0x+y=2,y=0,x=0 form a triangle in the first quadrant with vertices: (0,0), (2,0), (0,2).(0,0),\ (2,0),\ (0,2).(0,0), (2,0), (0,2).

So for x∈[0,2]x\in[0,2]x∈[0,2], the upper boundary of the triangle is y=2−x,y=2-x,y=2−x, and the lower boundary is y=0y=0y=0.

The required area is the area enclosed by this triangle and the curve y=f(x)=min⁡{x2+34, 1+[x]}.y=f(x)=\min\left\{x^2+\frac34,\ 1+[x]\right\}.y=f(x)=min{x2+43​, 1+[x]}.

Since the curve lies inside the triangle for part of the interval, the required area is A=∫02((2−x)−f(x)) dx.A=\int_0^2 \big((2-x)-f(x)\big)\,dx.A=∫02​((2−x)−f(x))dx. So first we must determine f(x)f(x)f(x) on [0,2][0,2][0,2].

  1. Evaluate f(x)f(x)f(x) piecewise using the floor function

Recall:

  • For 0≤x<10\le x<10≤x<1, [x]=0[x]=0[x]=0.
  • For 1≤x<21\le x<21≤x<2, [x]=1[x]=1[x]=1.
  • At x=2x=2x=2, [2]=2[2]=2[2]=2, but a single point does not affect area.

Thus:

For 0≤x<10\le x<10≤x<1

1+[x]=1.1+[x]=1.1+[x]=1. Also, x2+34∈[34,74).x^2+\frac34\in\left[\frac34,\frac74\right).x2+43​∈[43​,47​). We compare: x2+34≤1  ⟺  x2≤14  ⟺  0≤x≤12.x^2+\frac34\le 1 \iff x^2\le \frac14 \iff 0\le x\le \frac12.x2+43​≤1⟺x2≤41​⟺0≤x≤21​.

Hence on [0,1)[0,1)[0,1),

\begin{cases} x^2+\dfrac34, & 0\le x\le \dfrac12,\\[1mm] 1, & \dfrac12<x<1. \end{cases}$$ ### For $1\le x<2$ $$1+[x]=2.$$ But $$x^2+\frac34\ge 1+\frac34=\frac74<2 \quad \text{for }x=1,$$ and increases beyond $2$ later. Since we take minimum, compare: $$x^2+\frac34\le 2 \iff x^2\le \frac54 \iff x\le \frac{\sqrt5}{2}.$$ So on $[1,2)$, $$f(x)= \begin{cases} x^2+\dfrac34, & 1\le x\le \dfrac{\sqrt5}{2},\\[1mm] 2, & \dfrac{\sqrt5}{2}<x<2. \end{cases}$$ But notice: in the triangle, the upper boundary is $2-x$, which is at most $1$ for $x\ge 1$. Meanwhile on $[1,2)$, we have $$f(x)\ge \frac74>2-x.$$ So the curve lies **above** the line $y=2-x$ there, and contributes no enclosed strip of the form between $f(x)$ and $2-x$. Therefore the actual enclosed region occurs only where $$f(x)\le 2-x.$$ --- 3. **Find where the curve lies below the line $y=2-x$** We solve on the relevant pieces. ### Piece 1: $0\le x\le \frac12$ Here $$f(x)=x^2+\frac34.$$ Need $$x^2+\frac34\le 2-x.$$ That is, $$x^2+x-\frac54\le 0.$$ Multiply by 4: $$4x^2+4x-5\le 0.$$ Roots are $$x=\frac{-4\pm\sqrt{16+80}}{8}=\frac{-4\pm4\sqrt6}{8}=\frac{-1\pm\sqrt6}{2}.$$ Positive root is $$\alpha=\frac{\sqrt6-1}{2}\approx 0.7245.$$ Since on this piece $x\le \tfrac12<\alpha$, the inequality holds for all $x\in[0,\tfrac12]$. ### Piece 2: $\frac12<x<1$ Here $$f(x)=1.$$ Need $$1\le 2-x \iff x\le 1.$$ So this holds for all $x\in\left(\frac12,1\right].$$ Thus throughout $0\le x\le 1$, the curve lies below the line. For $x>1$, it does not. Hence $$A=\int_0^{1/2}\left[(2-x)-\left(x^2+\frac34\right)\right]dx+\int_{1/2}^{1}\left[(2-x)-1\right]dx.$$ --- 4. **Compute the integrals** First integral: $$\int_0^{1/2}\left(\frac54-x-x^2\right)dx.$$ Antiderivative: $$\frac54x-\frac{x^2}{2}-\frac{x^3}{3}.$$ Evaluate from $0$ to $\frac12$: $$\frac54\cdot\frac12-\frac{(1/2)^2}{2}-\frac{(1/2)^3}{3} =\frac58-\frac18-\frac1{24} =\frac12-\frac1{24} =\frac{11}{24}.$$ Second integral: $$\int_{1/2}^{1}(1-x)dx.$$ Antiderivative: $$x-\frac{x^2}{2}.$$ Evaluate: $$\left(1-\frac12\right)-\left(\frac12-\frac18\right) =\frac12-\frac38 =\frac18.$$ Therefore, $$A=\frac{11}{24}+\frac18=\frac{11}{24}+\frac{3}{24}=\frac{14}{24}=\frac{7}{12}.$$ So, $$12A=12\cdot\frac{7}{12}=7.$$ --- 5. **Compare with stored answer** My derived answer is $$\boxed{7}.$$ The stored correct answer is $17$, which does **not** match. So I disagree with the stored answer.
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