Math 31B ( 4C ) Discussion Section 3Practice Integration by Parts.The problem is to get a closed equation for I(a, b), a, b non-negative integers defined by:I(a, b) =Z10xa(1 − x)bdxa) We will first show I(a, b) = I(b, a)Notice that:I(a, b) =Z10xa(1 − x)bdx let u = 1 − x ⇒ du = −dx and x = 1 − u= −Z01(1 − u)aubdu=Z10(1 − u)aubdu=Z10ub(1 − u)adu= I(b, a)Where the final equation follows because it is the same as the original, integral, just with relabelling thevariable x as u.b) Now let’s find a formula for I(a, 0)I(a, 0) =Z10xadx =1a + 1And by part a), I(a, 0) = I(0, a) and therefore:I(0, a) =1a + 1c) Now, let’s show assuming a > 0: I(a, b) =ab + 1I(a − 1, b + 1)I(a, b) =Z10xa(1 − x)bdxNow let u =−1b + 1(1 − x)b+1, u0= (1 − x)b, v = xa, v0= axa−1Page 1 of 2Math 31B ( 4C ) Discussion Section 3I(a, b) =Z10xa(1 − x)bdx=Z10vu0= uv|10−Z10uv0=−1b + 1(1 − x)b+1xa10−Z10−1b + 1(1 − x)b+1axa−1dx= 0 − 0 +ab + 1Z10(1 − x)b+1xa−1dx =ab + 1I(a − 1, b + 1)d) Finishing upNow, take any I(a, b) by c),I(a, b) =ab + 1I(a − 1, b + 1)=a(a − 1)(b + 1)(b + 2)I(a − 2, b + 2)and repeating this for I(a − 3, b + 3), I(a − 4, b + 4) until the first argument of the I function is 0 andusing b),I(a, b) =a(a − 1) . . . 2 · 1(b + 1)(b + 2) . . . (b + a − 1)(b + a)I(0, b + a)=a!b!(b + a)!1b + a + 1=a!b!(b + a + 1)!So, finally, we get:I(a, b) =a!b!(b + a + 1)!.Page 2 of