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Definite Integration question

2022 · 26 Jun · Shift 1 · Q35
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  5. /2022 · 26 Jun · Shift 1 · Q35

Definite Integration question

2022 · 26 Jun · Shift 1 · Q35

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
Let f(x) = max {|x + 1|, |x + 2|, ....., |x + 5|}. Then ∫−60f(x)dx\int\limits_{ - 6}^0 {f(x)dx}−6∫0​f(x)dx is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 21

  1. We need to evaluate I=∫−60f(x) dx,I=\int_{-6}^{0} f(x)\,dx,I=∫−60​f(x)dx, where f(x)=max⁡{∣x+1∣,∣x+2∣,∣x+3∣,∣x+4∣,∣x+5∣}.f(x)=\max\{ |x+1|,|x+2|,|x+3|,|x+4|,|x+5|\}.f(x)=max{∣x+1∣,∣x+2∣,∣x+3∣,∣x+4∣,∣x+5∣}.

  2. Interpret the expression.

The numbers −1,−2,−3,−4,−5-1,-2,-3,-4,-5−1,−2,−3,−4,−5 are fixed points on the number line. For a given xxx, the quantities ∣x+1∣,∣x+2∣,…,∣x+5∣|x+1|,|x+2|,\dots,|x+5|∣x+1∣,∣x+2∣,…,∣x+5∣ are the distances from xxx to these points.

So f(x)f(x)f(x) is the maximum of these distances, i.e. the distance from xxx to the farthest point among −1,−2,−3,−4,−5-1,-2,-3,-4,-5−1,−2,−3,−4,−5.

Since the set of points ranges from −5-5−5 to −1-1−1, the farthest point from xxx will always be one of the endpoints: either −5-5−5 or −1-1−1. Thus f(x)=max⁡{∣x+5∣,∣x+1∣}.f(x)=\max\{|x+5|,|x+1|\}.f(x)=max{∣x+5∣,∣x+1∣}.

  1. Determine where each endpoint is farther.

We compare ∣x+5∣|x+5|∣x+5∣ and ∣x+1∣|x+1|∣x+1∣. They are equal when ∣x+5∣=∣x+1∣.|x+5|=|x+1|.∣x+5∣=∣x+1∣. Squaring or using midpoint idea, x=−3.x=-3.x=−3.

  • For x≤−3x\le -3x≤−3, the point −1-1−1 is farther, so f(x)=∣x+1∣.f(x)=|x+1|.f(x)=∣x+1∣.
  • For x≥−3x\ge -3x≥−3, the point −5-5−5 is farther, so f(x)=∣x+5∣.f(x)=|x+5|.f(x)=∣x+5∣.

Now on the interval [−6,0][-6,0][−6,0]:

  • for x∈[−6,−3]x\in[-6,-3]x∈[−6,−3], we have x+1≤0x+1\le 0x+1≤0, hence ∣x+1∣=−(x+1)=−x−1;|x+1|=-(x+1)=-x-1;∣x+1∣=−(x+1)=−x−1;
  • for x∈[−3,0]x\in[-3,0]x∈[−3,0], we have x+5≥0x+5\ge 0x+5≥0, hence ∣x+5∣=x+5.|x+5|=x+5.∣x+5∣=x+5.

Therefore,

\begin{cases} -x-1, & -6\le x\le -3,\\[4pt] x+5, & -3\le x\le 0. \end{cases}$$ 4. Split the integral. $$I=\int_{-6}^{-3}(-x-1)\,dx+\int_{-3}^{0}(x+5)\,dx.$$ 5. Compute each part. First part: $$\int(-x-1)dx=-\frac{x^2}{2}-x.$$ So, $$\int_{-6}^{-3}(-x-1)dx= \left[-\frac{x^2}{2}-x\right]_{-6}^{-3}.$$ At $x=-3$: $$-\frac{9}{2}+3=-\frac{3}{2}.$$ At $x=-6$: $$-\frac{36}{2}+6=-18+6=-12.$$ Hence, $$\int_{-6}^{-3}(-x-1)dx=-\frac{3}{2}-(-12)=\frac{21}{2}.$$ Second part: $$\int(x+5)dx=\frac{x^2}{2}+5x.$$ So, $$\int_{-3}^{0}(x+5)dx= \left[\frac{x^2}{2}+5x\right]_{-3}^{0}.$$ At $x=0$: $0$. At $x=-3$: $$\frac{9}{2}-15=-\frac{21}{2}.$$ Hence, $$\int_{-3}^{0}(x+5)dx=0-\left(-\frac{21}{2}\right)=\frac{21}{2}.$$ 6. Add them. $$I=\frac{21}{2}+\frac{21}{2}=21.$$ Therefore, $$\boxed{21}.$$
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