An engineer is interested in reducing the drying time of a particular paint. There are two formulations of the paint that are tested: formulation 1 is the typical one, and formulation 2 has a new ingredient that will reduce the drying time. It is known
that the standard deviation of drying time is 8 minutes, and this should be unaffected by the addition of the new ingredient.
10 items are painted with formulation 1, and another 10 items are painted with second formulation; the 20 items are painted
in random order. The two sample average drying times are 121 minutes for the first formulation and 112 minutes for the
second, respectively.
What conclusions can he make about the effectiveness of the new ingredient?​

A) y=x⁵-5x⁴+5x³+1       [-1;2]
y`=5x⁴-20x³+15x²=0
5*x²*(x²-4x+3)=0  |÷5
x²*(x-3)*(x-1)=0
x₁=0   x₂=1   x₃=3∉
y(-1)=(-1)⁵-5*(-1)⁴+5*(-1)³+1=-1-5*-5+1=-10=ymin
y(0)=0⁵-5*0⁴+5*0³+1=1
y(1)=1⁵-5*1⁴+5*1³+1=1-5+5+1=2=ymax
y(2)=2⁵-5*2⁴+5*2³+1=32-80+40+1=-7.
в) ₋₂∫⁰x²*e⁻ˣ/²dx  интегрируем по частям:
v=x²           dt=e⁻ˣ/²dx
dv=2xdx      t=-2*e⁻ˣ/²
∫vdt=vt-∫tdv=x²*(-2*e⁻ˣ/²)-∫(-2*e⁻ˣ/²*2x)dx=-2*x²*e⁻ˣ/²+4*∫(x*e⁻ˣ/²)dx
∫(x*e⁻ˣ/²)dx интегрируем по частям:
v=x        dt=e⁻ˣ/²dx
dv=dx     t=-2*e⁻ˣ/²
∫vdt=vt-₋∫(-2*e⁻ˣ/²)dx=x*(-2*e⁻ˣ/²)+2*∫(e⁻ˣ/²))dx=-2*x*e⁻ˣ/²-4*e⁻ˣ/²  ⇒
-2*x²*e⁻ˣ/²+4(-2*x*e⁻ˣ/²-4*e⁻ˣ/²)=-2*e⁻ˣ/²*(x²+4x+8) |⁰₋₂=
-2*e⁰*(0+0+8)-4*(-2*e¹*((-2)²+4*(-2)+8))=-16-(-2*e*(4-8+8))=
=-16+2*e*4=-16+8*e=8*(e-2)≈5,75.
u) y=x²   y=2-x²
x²=2-x²
2x²=2  |÷2
x²=1
x₁=-1    x₂=1
S=₋₁∫¹(2-x²-x²)dx=₋₁∫¹(2-2x²)dx=2*₋₁∫¹(1-x²)dx=2*x-2*x³/3) |¹₋₁=
=2*1-2*1³/3)-(2*(-1)-2*(-1)³/3))=2-2/3-(-2+2/3)=1¹/₃+1¹/₃=2²/₃=8/3.
ответ: S=8/3=2,67 кв. ед.

A) y=x⁵-5x⁴+5x³+1       [-1;2]
y`=5x⁴-20x³+15x²=0
5*x²*(x²-4x+3)=0  |÷5
x²*(x-3)*(x-1)=0
x₁=0   x₂=1   x₃=3∉
y(-1)=(-1)⁵-5*(-1)⁴+5*(-1)³+1=-1-5*-5+1=-10=ymin
y(0)=0⁵-5*0⁴+5*0³+1=1
y(1)=1⁵-5*1⁴+5*1³+1=1-5+5+1=2=ymax
y(2)=2⁵-5*2⁴+5*2³+1=32-80+40+1=-7.
в) ₋₂∫⁰x²*e⁻ˣ/²dx  интегрируем по частям:
v=x²           dt=e⁻ˣ/²dx
dv=2xdx      t=-2*e⁻ˣ/²
∫vdt=vt-∫tdv=x²*(-2*e⁻ˣ/²)-∫(-2*e⁻ˣ/²*2x)dx=-2*x²*e⁻ˣ/²+4*∫(x*e⁻ˣ/²)dx
∫(x*e⁻ˣ/²)dx интегрируем по частям:
v=x        dt=e⁻ˣ/²dx
dv=dx     t=-2*e⁻ˣ/²
∫vdt=vt-₋∫(-2*e⁻ˣ/²)dx=x*(-2*e⁻ˣ/²)+2*∫(e⁻ˣ/²))dx=-2*x*e⁻ˣ/²-4*e⁻ˣ/²  ⇒
-2*x²*e⁻ˣ/²+4(-2*x*e⁻ˣ/²-4*e⁻ˣ/²)=-2*e⁻ˣ/²*(x²+4x+8) |⁰₋₂=
-2*e⁰*(0+0+8)-4*(-2*e¹*((-2)²+4*(-2)+8))=-16-(-2*e*(4-8+8))=
=-16+2*e*4=-16+8*e=8*(e-2)≈5,75.
u) y=x²   y=2-x²
x²=2-x²
2x²=2  |÷2
x²=1
x₁=-1    x₂=1
S=₋₁∫¹(2-x²-x²)dx=₋₁∫¹(2-2x²)dx=2*₋₁∫¹(1-x²)dx=2*x-2*x³/3) |¹₋₁=
=2*1-2*1³/3)-(2*(-1)-2*(-1)³/3))=2-2/3-(-2+2/3)=1¹/₃+1¹/₃=2²/₃=8/3.
ответ: S=8/3=2,67 кв. ед.