JEE Main 2020 — Mathematics Calculus

$$I=\frac{1}{(a+b)}\int _{a}^{b}x[f(x)+f(x+1)]dx\ldots (1)$$I=(a+b)1​∫ab​x[f(x)+f(x+1)]dx…(1)

$$x\rightarrow a+b-x$$x→a+b−x

$$I=\frac{1}{(a+b)}\int _{a}^{b}(a+b-x)[f(a+b-x)+(a+b+1-x)]dx$$I=(a+b)1​∫ab​(a+b−x)[f(a+b−x)+(a+b+1−x)]dx

$$I=\frac{1}{(a+b)}\int _{a}^{b}a(a+b-x)[f(x+1)+f(x)dx]\ldots (2)$$I=(a+b)1​∫ab​a(a+b−x)[f(x+1)+f(x)dx]…(2)

$\because$∵ Keeping $x\rightarrow x+1$x→x+1 in given relation

$$(1)+(2)$$(1)+(2)

$$2I=\int _{a}^{b}[f(x+1)+f(x)]dx$$2I=∫ab​[f(x+1)+f(x)]dx

$$2I=\int _{a}^{b}f(x+1)dx+\int _{a}^{b}f(x)dx$$2I=∫ab​f(x+1)dx+∫ab​f(x)dx

$$\int _{a}^{b}f(a+b+1-x)dx+\int _{a}^{b}f(x)dx$$∫ab​f(a+b+1−x)dx+∫ab​f(x)dx

$$2I=2\int _{a}^{b}f(x)dx$$2I=2∫ab​f(x)dx

$$I=\int _{a}^{b}f(x)dx$$I=∫ab​f(x)dx

Put $\displaystyle x=t+1\Rightarrow I=\int _{a-1}^{b-1}f(t+1)dt=\int _{a-1}^{b-1}f(x+1)dx$x=t+1⇒I=∫a−1b−1​f(t+1)dt=∫a−1b−1​f(x+1)dx