JEE Main 2024 — Mathematics Calculus

Let,

$$I={\int }_{0}^{9}[\sqrt{\frac{10x}{x+1}}dx]$$I=∫09​[x+110x​​dx]

The square of integers are $0,1,4,9,...$0,1,4,9,...

$$\Rightarrow 0=\frac{10x}{x+1}\Rightarrow x=0...(i)$$⇒0=x+110x​⇒x=0...(i)

$$\Rightarrow 1=\frac{10x}{x+1}\Rightarrow x+1=10x$$⇒1=x+110x​⇒x+1=10x

$$\Rightarrow x=\frac{1}{9}...(ii)$$⇒x=91​...(ii)

$$\Rightarrow 4=\frac{10x}{x+1}\Rightarrow 4x+4=10x$$⇒4=x+110x​⇒4x+4=10x

$$\Rightarrow x=\frac{2}{3}...(iii)$$⇒x=32​...(iii)

$$\Rightarrow 9=\frac{10x}{x+1}\Rightarrow 9x+9=10x$$⇒9=x+110x​⇒9x+9=10x

$$\Rightarrow x=9...(iv)$$⇒x=9...(iv)

Now, from the above equations using the law $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{c}f(x)dx+{\int }_{c}^{d}f(x)dx+{\int }_{d}^{b}f(x)dx$∫ab​f(x)dx=∫ac​f(x)dx+∫cd​f(x)dx+∫db​f(x)dx

$$\Rightarrow I={\int }_{0}^{\frac{1}{9}}(0)dx+{\int }_{\frac{1}{9}}^{\frac{2}{3}}(1)dx+{\int }_{\frac{2}{3}}^{9}(2)dx$$⇒I=∫091​​(0)dx+∫91​32​​(1)dx+∫32​9​(2)dx

$$\Rightarrow I=\frac{2}{3}-\frac{1}{9}+2(9-\frac{2}{3})$$⇒I=32​−91​+2(9−32​)

$$\Rightarrow I=\frac{155}{9}$$⇒I=9155​

$$\Rightarrow 9I=155$$⇒9I=155