y = x√(2ax - x²)

Method is to integrate by substitution.

Let x = a sin t + a
dx/dt = a cos t
dx = a cos t dt

when x = 0, a sin t + a = 0
a sin t = - a
sin t = - 1
t = -π/2


when x = a, a sin t + a = a
a sin t = 0
sin t = 0
t = 0


∫ x√(2ax - x²) dx = 888 becomes

∫ (a sin t + a)√[(2a(a sin t + a) - (a sin t + a)²] (a cos t) dt = 888

(You can skip some steps below if needed)

∫ (a sin t + a)√[(2a² sin t + 2a² - a² sin² t - 2a² sin t - a²] (a cos t) dt = 888

∫ (a sin t + a)√(a² - a² sin² t) (a cos t) dt= 888

∫ (a sin t + a)√(a² cos² t) (a cos t) dt = 888

∫ (a sin t + a)(a cos t)(a cos t) dt = 888

∫ (a sin t + a)(a² cos² t) dt = 888

∫ (a³ sin t cos² t + a³ cos² t)dt = 888

∫ (a³ sin t cos² t + a³( 1 + cos 2t)/2 ) dt = 888

[ -⅓a³ cos³t + ½a³ (t + ½sin 2t) ] (-π,0) = 888

[ -⅓a³ cos³ (0) + ½a³ (0 + ½sin 2(0)] - [-⅓a³ cos³ (-π/2) + ½a³ (-π/2 + ½sin2(-π/2)) ] = 888

-⅓ a³ - (½a³(-π/2))= 888

a³(¼π - ⅓) = 888

a³ = 888/(¼π - ⅓)

a = ³√[888/(¼π - ⅓)]

a ≈ 12.5238372
= 12.52 (2d.p)

Alternatively, you can do double substitution.

y = x√(2ax - x²)

① Complete the square first by adding a² term inside.

y = x√(a² - x² + 2ax - a²)
y = x√(a² - (x² - 2ax + a²))
y = x√(a² - (x - a)²)

Let u = x - a → x = u + a
du/dx = 1
du = dx


∫ x√(a² - (x - a)²) dx becomes

∫ (u + a)√(a² - u²) du


② Perform substitution again.

Let u = a sin t

du/dt = a cos t

du = a cos t dt


Then ∫ (u + a)√(a² - u²) du becomes

∫ (a sin t + a)√(a² - a²sin²t) (a cos t ) dt


Change the bounds in the same manner as the previous comment and solve in the same way accordingly.