Difficulties with Integrals...very hard?

Question

Evaluate (int 4/1) {x * sqrtx (x-1) dx}


If anyone can help with this question that's great...
I'm using the (int) f(x) dx = F(4) - F(1).

Answer 1

I take it you mean Integ_4_1 [ x * √x (x-1) dx ]
and the √ only applies to the (x) not the (x-1) bracket?
Please clarify?

If it's √((x)(x-1)) you need a substitution.
Complete the square on x and that will introduce a constant inside the radical:
√((x)(x-1)) = √(x² -x)
= √(x² -2(1/2)x + (1/2)² - (1/2)²)
= √((x-1/2)² - (1/2)²)
= (1/2) √((2x-1)² - 1²)

This is a standard form √(u² - 1), so let u = sec t
then √(u² - 1) = tan t
du = -sect tant t dt
=> dt = -du / u√(u² - 1)

sec t = (2x-1)
=> 2 dx = -sec t tan t dt
√((2x-1)² - 1²) = tan t

Integ_4_1 [ x * √(x)(x-1) ] dx
= Integ_4_1 [ x (1/2) √((2x-1)² - 1²) dx ]
= Integ [ ((1+ sec t)/2) (1/2) (tan t) (- (1/2)sec t tan t dt) ]
= Integ -1/8[ ((1+ sec t) sec t tan² t dt ]
= Integ -1/8[ ((1+ sec t) sec t tan² t dt ]
etc... multiply it out, just a bunch of standard forms.

If it's just √(x), you have a sum of separate powers of x and can directly integrate that:
x * √(x) * (x-1)
= (x² -x) * √(x)
= x^2.5 - x^1.5
and Integ_4_1 [ x^2.5 - x^1.5 dx ] is easy

Answer 2

♦ thus y(x)*dx = dx*x*√(x-1);
t=√(x-1), x=t^2+1, dx=2t*dt, t1=√(1-1)=0; t2=√(4-1)=√3;
♥ y(t)*dt =2t*dt*( t^2+1)*t = 2t^4*dt +2t^2*dt;
go on!