If \int\limits_ - a^a ( | x | + | x - 2 | ) dx = 22 , (a > 2) and [x] denotes the greatest integer \le x, then \int\limits_ - a^a ( x + [ x ] ) dx is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2021 · 24 Feb · Shift 1 · Q42 with Solution

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Correct answer: -3

$\int_{-a}^{a}\left(|x|+|x-2|\right)dx=22, \qquad a>2.$

Since $a>2$ , over the interval $[-a,a]$ we split according to the critical points $0$ and $2$ .

Hence

=\int_{-a}^{0}(2-2x)dx+\int_{0}^{2}2\,dx+\int_{2}^{a}(2x-2)dx.$$ Now compute each part: $$\int_{-a}^{0}(2-2x)dx=\left[2x-x^2\right]_{-a}^{0}=0-(-2a-a^2)=a^2+2a,$$ $$\int_{0}^{2}2\,dx=4,$$ $$\int_{2}^{a}(2x-2)dx=\left[x^2-2x\right]_{2}^{a}=a^2-2a.$$ Therefore, $$a^2+2a+4+a^2-2a=22$$ $$2a^2+4=22$$ $$2a^2=18$$ $$a^2=9.$$ Since $a>2$, we get $$a=3.$$ --- 2. Now evaluate $$I=\int_{-a}^{a}(x+[x])dx=\int_{-3}^{3}(x+[x])dx.$$ So $$I=\int_{-3}^{3}x\,dx+\int_{-3}^{3}[x]dx.$$ The first integral is zero because $x$ is odd over $[-3,3]$: $$\int_{-3}^{3}x\,dx=0.$$ Thus $$I=\int_{-3}^{3}[x]dx.$$ --- 3. Evaluate $\int_{-3}^{3}[x]dx$ by splitting into unit intervals. Recall: - On $[-3,-2)$, $[x]=-3$ - On $[-2,-1)$, $[x]=-2$ - On $[-1,0)$, $[x]=-1$ - On $[0,1)$, $[x]=0$ - On $[1,2)$, $[x]=1$ - On $[2,3)$, $[x]=2$ (The values at isolated endpoints do not affect the integral.) Hence $$\int_{-3}^{3}[x]dx =\int_{-3}^{-2}(-3)dx+\int_{-2}^{-1}(-2)dx+\int_{-1}^{0}(-1)dx+ \int_{0}^{1}0\,dx+\int_{1}^{2}1\,dx+\int_{2}^{3}2\,dx.$$ So $$= (-3)(1)+(-2)(1)+(-1)(1)+0(1)+1(1)+2(1)$$ $$= -3-2-1+0+1+2=-3.$$ Therefore, $$\int_{-3}^{3}(x+[x])dx=-3.$$ --- 4. Final answer: $$\boxed{-3}$$ The stored correct answer is $3$, but the correct evaluated integral is $-3$.

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Correct answer: -3

$\int_{-a}^{a}\left(|x|+|x-2|\right)dx=22, \qquad a>2.$

Since $a>2$ , over the interval $[-a,a]$ we split according to the critical points $0$ and $2$ .

Hence

=\int_{-a}^{0}(2-2x)dx+\int_{0}^{2}2\,dx+\int_{2}^{a}(2x-2)dx.$$ Now compute each part: $$\int_{-a}^{0}(2-2x)dx=\left[2x-x^2\right]_{-a}^{0}=0-(-2a-a^2)=a^2+2a,$$ $$\int_{0}^{2}2\,dx=4,$$ $$\int_{2}^{a}(2x-2)dx=\left[x^2-2x\right]_{2}^{a}=a^2-2a.$$ Therefore, $$a^2+2a+4+a^2-2a=22$$ $$2a^2+4=22$$ $$2a^2=18$$ $$a^2=9.$$ Since $a>2$, we get $$a=3.$$ --- 2. Now evaluate $$I=\int_{-a}^{a}(x+[x])dx=\int_{-3}^{3}(x+[x])dx.$$ So $$I=\int_{-3}^{3}x\,dx+\int_{-3}^{3}[x]dx.$$ The first integral is zero because $x$ is odd over $[-3,3]$: $$\int_{-3}^{3}x\,dx=0.$$ Thus $$I=\int_{-3}^{3}[x]dx.$$ --- 3. Evaluate $\int_{-3}^{3}[x]dx$ by splitting into unit intervals. Recall: - On $[-3,-2)$, $[x]=-3$ - On $[-2,-1)$, $[x]=-2$ - On $[-1,0)$, $[x]=-1$ - On $[0,1)$, $[x]=0$ - On $[1,2)$, $[x]=1$ - On $[2,3)$, $[x]=2$ (The values at isolated endpoints do not affect the integral.) Hence $$\int_{-3}^{3}[x]dx =\int_{-3}^{-2}(-3)dx+\int_{-2}^{-1}(-2)dx+\int_{-1}^{0}(-1)dx+ \int_{0}^{1}0\,dx+\int_{1}^{2}1\,dx+\int_{2}^{3}2\,dx.$$ So $$= (-3)(1)+(-2)(1)+(-1)(1)+0(1)+1(1)+2(1)$$ $$= -3-2-1+0+1+2=-3.$$ Therefore, $$\int_{-3}^{3}(x+[x])dx=-3.$$ --- 4. Final answer: $$\boxed{-3}$$ The stored correct answer is $3$, but the correct evaluated integral is $-3$.