The curve y=ax2+bx passes through (1,2) so 2‌‌=a(1)2+b(1) a+b‌‌=2‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) Now the area under the curve is,
6
∫
0
(ax2+bx)dx=108 [‌
ax3
3
+‌
bx2
2
]06‌‌=108 72a+18b‌‌=108 4a+b‌‌=6‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) From equation (I) and (II), a=‌