Answer:
The integrals was calculated.
Step-by-step explanation:
We calculate integrals, and we get:
1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}
2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}
3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}
4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}
5) \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=
=\frac{3π+8}{64}
6) ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x
7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}
8) ∫ tan^5 (x) sec(x) dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x
Answer:
h=1/2fg
Step-by-step explanation:
Solve for x, h=1/2fg
It is true for all x; h=1/2fg
h=1/2fg
Both sides are equal
It is true for all x; h=1/2fg
Answer:3/4 inch
Step-by-step explanation:
Use the substitution method:
2x - 6y = -12
x - 2y = -8
Then x = 2y - 8
Substitute in the first equation:
2(2y - 8) - 6y = -12
4y - 16 - 6y = -12
-2y = 4
y = -2
Now substitute y in one of the two equations given you prefer:
For example x-2*(-2) = -8
x + 4 = -8
x = -12
The solutions are x = -12 and y = -2
