# Applications of Definite Integrals

##### 313 words — categories: calculus

## Total Profit

A company finds that the marginal profit, in dollars, from drilling a well that is \(x\) feet deep is given by

\[P^\prime(x) = \sqrt[5]{x}\] Find the profit when a well 250 ft deep is drilled.

```
import sympy as sp
from sympy.abc import x
import matplotlib.pyplot as plt
import numpy as np
marginal_profit = sp.root(x, 5)
marginal_profit
```

`## x**(1/5)`

```
profit = sp.integrate(marginal_profit, x)
profit
```

`## 5*x**(6/5)/6`

```
p = sp.lambdify(x, profit)
p(250)
```

`## 628.5600350567877`

`sp.integrate(marginal_profit, (x, 0, 250)).evalf()`

`## 628.560035056788`

```
x_values = np.linspace(0, 300)
plt.plot(x_values, p(x_values))
plt.fill_between(x_values, p(x_values), where=[x <= 250 for x in x_values], color="green", alpha=0.3)
```

## Increasing Total Profit

Laso Industries finds that the marginal profit, in dollars from the sale of \(x\) digital control boards is given by

\[P^\prime(x) = 2.6x^{0.1}\] A customer orders 1200 digital control boards, and later increases the order to 1500. Find the extra profit resulting from the increase in order size.

```
marginal_profit = (2.6 * x)**0.1
profit = sp.integrate(marginal_profit, x)
profit
```

`## 1.00024099373274*x**1.1`

```
second_purchase = sp.integrate(marginal_profit, (x, 0, 1500))
second_purchase
```

`## 3117.48975261839`

```
first_purchase = sp.integrate(marginal_profit, (x, 0, 1200))
first_purchase
```

`## 2438.95630774585`

```
extra_profit = second_purchase - first_purchase
extra_profit
```

`## 678.533444872540`

```
x_values = np.linspace(0, 2000)
p = sp.lambdify(x, profit)
plt.plot(x_values, p(x_values))
plt.fill_between(x_values, p(x_values), where=[x <= 1200 for x in x_values], color="green", alpha=0.3)
plt.fill_between(x_values, p(x_values), where=[(x <= 1500) and (x > 1200) for x in x_values], color="red", alpha=0.3)
```

## Accumulated Sales

Ragss, Ltd. estimates that its sales will grow continuously at a rate given by

\[S^\prime(t) = 10e^t\] where \(S^\prime(t)\) is the rate at which sales are increasing, in dollars per day, on day \(t\).

- Find the accumulated sales for the first 5 days.
- Find the sales from the 2nd day through the 5th day.

```
from math import e
from sympy.abc import t
marginal_sales = (10 * e)**t
marginal_sales
```

`## 27.1828182845905**t`

```
sales = sp.integrate(marginal_sales, t)
sales
```

`## 0.302793106564114*27.1828182845905**t`

`sp.integrate(marginal_sales, (t, 0, 5)).evalf()`

`## 4493847.84717322`

`sp.integrate(marginal_sales, (t, 2, 5)).evalf()`

`## 4493624.41444125`