Given curve y=a√x+bx passes through (1,2). The area formed by curve line x=4 and X-axis is 8 sq units. ∵ Since, the curves y=a√x+bx passes through (1,2). ∴2=a√1+b⋅1\a+b=2...(i) Now, the area enclosed by the curve line x=4 and X-axis =
4
∫
0
ydx=
4
∫
0
(a√x+bx)dx =[a⋅
x
3
2
3
2
+b⋅
x2
2
]04 =
2a
3
⋅4
3
2
+
b
2
⋅42=
16a
3
+8b From the problem
16a
3
+8b=8 ∴2a+3b=3...(ii) From Eqs. (i) and (ii), we get a=3,b=−1