If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then 1 \over a + b\int_a^b x( f(x) + f(x + 1) ) dx is equal to:

Definite Integration PYQ — JEE Main 2020 · 7 Jan · Shift 1 · Q38 with Solution

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Correct answer: D

Let $I=\frac{1}{a+b}\int_a^b x\big(f(x)+f(x+1)\big)\,dx.$ We use the given symmetry $f(a+b+1-x)=f(x)\quad \text{for all }x.$
Split the integral: $I=\frac{1}{a+b}\left(\int_a^b x f(x)\,dx+\int_a^b x f(x+1)\,dx\right).$ Let $I_1=\int_a^b x f(x)\,dx,\qquad I_2=\int_a^b x f(x+1)\,dx.$ So, $I=\frac{I_1+I_2}{a+b}.$
Transform $I_1$ using the substitution $u=a+b+1-x \quad\Rightarrow\quad x=a+b+1-u,\quad dx=-du.$ When $x=a$ , $u=b+1$ ; when $x=b$ , $u=a+1$ . Thus

=\int_{a+1}^{b+1}(a+b+1-u)f(u)\,du,$$ using $f(a+b+1-u)=f(u)$. Now put $u=t+1$, so $t=u-1$, $du=dt$. Then as $u$ goes from $a+1$ to $b+1$, $t$ goes from $a$ to $b$: $$I_1=\int_a^b (a+b-t)f(t+1)\,dt.$$ Renaming $t$ as $x$, $$I_1=\int_a^b (a+b-x)f(x+1)\,dx.$$ 4. Therefore, $$I_1+I_2=\int_a^b (a+b-x)f(x+1)\,dx+\int_a^b x f(x+1)\,dx.$$ Combine them: $$I_1+I_2=\int_a^b (a+b)f(x+1)\,dx=(a+b)\int_a^b f(x+1)\,dx.$$ Hence $$I=\frac{1}{a+b}(I_1+I_2)=\int_a^b f(x+1)\,dx.$$ 5. Now shift the variable: let $u=x-1$, so $x=u+1$, $dx=du$. When $x=a$, $u=a-1$; when $x=b$, $u=b-1$. Thus $$\int_a^b f(x+1)\,dx=\int_{a-1}^{b-1} f(u+2)\,du,$$ which is not directly among the options. Instead, use the standard substitution $t=x-1$ in the option expressions: $$\int_a^b f(x+1)\,dx=\int_{a+1}^{b+1} f(t)\,dt.$$ This matches option D. Also, $$\int_{a-1}^{b-1} f(x+1)\,dx=\int_a^b f(t)\,dt,$$ which is generally not the same as $\int_a^b f(x+1)\,dx$. 6. So the correct value is $$\boxed{\int_{a+1}^{b+1} f(x)\,dx}.$$ Therefore, the correct option is **D**.

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Correct answer: D

Let $I=\frac{1}{a+b}\int_a^b x\big(f(x)+f(x+1)\big)\,dx.$ We use the given symmetry $f(a+b+1-x)=f(x)\quad \text{for all }x.$
Split the integral: $I=\frac{1}{a+b}\left(\int_a^b x f(x)\,dx+\int_a^b x f(x+1)\,dx\right).$ Let $I_1=\int_a^b x f(x)\,dx,\qquad I_2=\int_a^b x f(x+1)\,dx.$ So, $I=\frac{I_1+I_2}{a+b}.$
Transform $I_1$ using the substitution $u=a+b+1-x \quad\Rightarrow\quad x=a+b+1-u,\quad dx=-du.$ When $x=a$ , $u=b+1$ ; when $x=b$ , $u=a+1$ . Thus

=\int_{a+1}^{b+1}(a+b+1-u)f(u)\,du,$$ using $f(a+b+1-u)=f(u)$. Now put $u=t+1$, so $t=u-1$, $du=dt$. Then as $u$ goes from $a+1$ to $b+1$, $t$ goes from $a$ to $b$: $$I_1=\int_a^b (a+b-t)f(t+1)\,dt.$$ Renaming $t$ as $x$, $$I_1=\int_a^b (a+b-x)f(x+1)\,dx.$$ 4. Therefore, $$I_1+I_2=\int_a^b (a+b-x)f(x+1)\,dx+\int_a^b x f(x+1)\,dx.$$ Combine them: $$I_1+I_2=\int_a^b (a+b)f(x+1)\,dx=(a+b)\int_a^b f(x+1)\,dx.$$ Hence $$I=\frac{1}{a+b}(I_1+I_2)=\int_a^b f(x+1)\,dx.$$ 5. Now shift the variable: let $u=x-1$, so $x=u+1$, $dx=du$. When $x=a$, $u=a-1$; when $x=b$, $u=b-1$. Thus $$\int_a^b f(x+1)\,dx=\int_{a-1}^{b-1} f(u+2)\,du,$$ which is not directly among the options. Instead, use the standard substitution $t=x-1$ in the option expressions: $$\int_a^b f(x+1)\,dx=\int_{a+1}^{b+1} f(t)\,dt.$$ This matches option D. Also, $$\int_{a-1}^{b-1} f(x+1)\,dx=\int_a^b f(t)\,dt,$$ which is generally not the same as $\int_a^b f(x+1)\,dx$. 6. So the correct value is $$\boxed{\int_{a+1}^{b+1} f(x)\,dx}.$$ Therefore, the correct option is **D**.