\(\int4e^{-7x}dx\)
u = -7x
du = -7dx
\(\int4e^{-7x}dx = \int \frac{-7x4}{-7} e^{-7x} dx\)
\(=\int\frac{-4}{7} e^{u} du\)
\(=\frac{-4}{7}e^u + C\)
\(=-\frac{4}{7}e^{-7x} + C\)
\(~\)
\(\frac{dN}{dt} = N'(t)\)
\(=-\frac{3150}{t^4}-220\)
\(N(t) = \int(\frac{-3150}{t^4} -220)dt=\frac{1050}{t^3}-200t + C\)
\(N(1) = 6530\)
\(N(t) = \frac{1050}{t^3} -220t +C\)
\(= \frac{1050}{1^3}-220 * 1 + C = 6530\)
\(C = 6530 - 1050 +220\)
C = 5700
\(~\)
f(x) = 2x - 9
\(\int_{4.5}^{8.5} 2x = 9dx\)
\(=[x^2 - 9x]|^{8.5}_{4.5}\)
\(=(8.5^2 - 9(8.5))-(4.5^2 - 9(4.5))\)
Area of red rectangles = 16
\(~\)
\(y = x^2 - 2x - 2\), \(y = x + 2\)
# Create funtions
<- function(x){x^2 - (2*x) - 2}
y1 <- function(x){x + 2}
y2
# plot lines
curve(y1, from = -6, to = 6, col = 2)
curve(y2, from = -6, to = 6, add = TRUE, col = 4)
integrate(y1, lower = -1, upper = 4)
## -3.333333 with absolute error < 1.2e-13
integrate(y2, lower = -1, upper = 4)
## 17.5 with absolute error < 1.9e-13
= 17.5 - (-3.33)
total_area total_area
## [1] 20.83
\(~\)
n = number orders per year
x = number of flat irons to order per year
x / 2 = average inventory level
Storage cost = 3.75 * (x / 2) Storage cost = 1.875x
Order cost = 8.25 * 110 / x Order cost = 907.5 / x
Inventory cost = 1.875x + (907.5 /x)
Therefore:
1.875 - 907.5 * x - 2 = 0
x = \(\sqrt484\)
x = 22
Orders per year = 110 / n = 22
n = 5
\(~\)
\(\int ln(9x) * x^6 dx\)
u = In(9x) therefore du = \(\frac{1}{x}dx\)
v = \(\frac{1}{7}x^7\) therefore \(dv = x^6dx\)
With integration we get \(\int uv' = uv = \int u'v\)
\(\int ln(9x)x^6dx = ln(9x)\frac{1}{7}x^7 - \int\frac{1}{x}\frac{1}{7}x^7dx\)
\(= \frac{1}{7}ln(9x)x^7 - \frac{1}{49}x^7 +C\)
\((\frac{1}{7}ln(9x)x^7 - \frac{1}{49}x^7 +C\)
\(~\)
\(f(x) = \frac{1}{6x}\)
\(\int_{1}^{e^6} \frac{1}{6x}dx = \frac{1}{6}int_{1}^{e^6}\frac{1}{x}dx\)
\(= \frac{1}{6}(ln(e^6) - ln(1))\)
= 1 therefore f(x) is a probability density function on the interval \([1, e^6]\)