If (a,b) be the orthocentre of the triangle whose vertices are (1,2),(2,3) and (3,1), and I_1=∫ _a^b xsin(4x-x^2) dx,I_2=∫ _a^b sin(4x-x^2) dx , then…

Equation of $\mathrm{CE}$CE is given by,

$$\Rightarrow (y-1)=\frac{-1}{\frac{3-2}{2-1}}(x-3)$$⇒(y−1)=2−13−2​−1​(x−3)

$$\Rightarrow (y-1)=-(x-3)$$⇒(y−1)=−(x−3)

$$\Rightarrow x+y=4...(i)$$⇒x+y=4...(i)

Similarly, equation of $AD$AD is given by,

$$\Rightarrow (y-2)=\frac{-1}{\frac{3-1}{2-3}}(x-1)$$⇒(y−2)=2−33−1​−1​(x−1)

$$\Rightarrow (y-2)=\frac{1}{2}(x-1)$$⇒(y−2)=21​(x−1)

$$\Rightarrow 2y-4=x-1$$⇒2y−4=x−1

$$\Rightarrow x-2y+3=0$$⇒x−2y+3=0

Using $(i)$(i)

$$\Rightarrow 4-y-2y+3=0$$⇒4−y−2y+3=0

$$\Rightarrow y=\frac{7}{3}$$⇒y=37​

$$\Rightarrow x=4-\frac{7}{3}$$⇒x=4−37​

$$\Rightarrow x=\frac{5}{3}$$⇒x=35​

So, Orthocentre is $H\equiv (a,b)=(\frac{5}{3},\frac{7}{3})$H≡(a,b)=(35​,37​).

Orthocentre lies on the line $x+y=4$x+y=4

So, $a+b=4$a+b=4

Now, $\displaystyle {I}_{1}={\int }_{a}^{b}x\mathrm{sin}(x(4-x))dx...(1)$I1​=∫ab​xsin(x(4−x))dx...(1)

$$\Rightarrow {I}_{1}={\int }_{a}^{b}(a+b-x)\mathrm{sin}((a+b-x)(4-(a+b-x)))dx$$⇒I1​=∫ab​(a+b−x)sin((a+b−x)(4−(a+b−x)))dx

$$\Rightarrow {I}_{1}={\int }_{a}^{b}(4-x)\mathrm{sin}(x(4-x))dx...(2)$$⇒I1​=∫ab​(4−x)sin(x(4−x))dx...(2)

Using, $(1)+(2)$(1)+(2)

$$\Rightarrow 2{I}_{1}={\int }_{a}^{b}4\mathrm{sin}(x(4-x))dx$$⇒2I1​=∫ab​4sin(x(4−x))dx

$$\Rightarrow 2{I}_{1}=4{I}_{2}$$⇒2I1​=4I2​

$$\Rightarrow {I}_{1}=2{I}_{2}$$⇒I1​=2I2​

$$\Rightarrow \frac{{I}_{1}}{{I}_{2}}=2$$⇒I2​I1​​=2

$$\Rightarrow \frac{36{I}_{1}}{{I}_{2}}=72$$⇒I2​36I1​​=72