a. 1^5 = 1 × 1 × 1 × 1 × 1 = 1
b. 2 × 2 × 2 × 2 × 2 = 2^5 = 32
c. 20^3 = 20 × 20 × 20 = 8.000
d. 7 × 7 × 7 × 7 = 7^4 = 2.401
Rather than carrying out IBP several times, let's establish a more general result. Let
One round of IBP, setting
gives
This is called a power-reduction formula. We could try solving for 
explicitly, but no need. 
is small enough to just expand 
as much as we need to.
Finally,
so we end up with
and the antiderivative is
X=width of a coccer pitch
2x-20=length of a soccer pitch
Area (rectangule)=length x width
We suggest this equation:
x(2x-20)=6000
2x²-20x=6000
2x²-20x-6000=0
x²-10x-3000=0
We solve this quadratic equation:
x=[10⁺₋√(100-4*1*-3000)]/2=[10⁺₋√(100+12000)]/2=
=(10⁺₋110)/2
we have two solutions:
x₁=(10-110)/2=-50, invalid solution.
x₂=(10+110)/2=60
x=60
2x-20=2(60)-20=120-20=100
Solution: the length is 100 m, and the width is 60 m.
To check:
Area=100 m*60 m=6000 m²
The twice of width is =2(60 m)=120 m,
20 m less than twice its width is: 120 m-20 m=100 m=the length.
The answer is -125. Get this by taking (-5)(-5)(-5)
