Can someone help check my work I solved but the answer is not the same

[frank glen]

At which part?

[leaden bolt]

, tex Let the line from O to A be $l$, the process of finding the shaded area includes solving $\int_{0}^{a} (y-l) dx$

[fiery frigate]

Lunar

[leaden bolt]

you forgot to square the 3 on the second line beginning from the blue colored pen

[leaden bolt]

, tex We can seperate the integral into two: $\int_{0}^{a} ydx -\int_{0}^{a} ldx$, but we know the area from 0 to $a$ under the line is just the area of a triangle: $\therefore \int_{0}^{a} ldx = \dfrac{1}{2}ay(a)$ and $y(a) = a^3 - 6a^3 + 9a^3$, then: $\int_{0}^{a} ldx = 2a^4$, thus the area is $A = \dfrac{11}{4}a^4 - 2a^4 = \dfrac{3}{4}a^4$

[fiery frigate]

Lunar