INTEGRATION Power Rule for Integration

INTEGRATION

Power Rule for Integration

Last week we learnt that the integral of a power function in the form 𝑥𝑎 , where 𝑥 is a variable is given as:

For more general power functions, say for instance,∫(𝑥 + 1)20𝑑𝑥, applying the rule above may not work because the (𝑥 + 1) is a function and not a variable. When this happens, one option is to transform the original integral into the form of the integral ∫ 𝑥𝑎𝑑𝑥 = 𝑥_𝑎+1𝑎+1+ 𝑐 , and work backwards after the integration.

Example.

Find ∫(𝑥 + 1)20𝑑𝑥

Solution:

Suppose we let 𝑢 = 𝑥 + 1. Then,

𝑑𝑢 𝑑𝑥 = 1 𝑑𝑢 = 𝑑𝑥

When all instances of 𝑥 are replaced with 𝑢, then ∫(𝑥 + 1)20𝑑𝑥 = ∫ 𝑢20_{𝑑𝑢 =}𝑢21 21 + 𝑐 After the integration, we can replace back all instances of 𝑢 with 𝑥.

⇒ ∫(𝑥 + 1)20_{𝑑𝑥 =}(𝑥+1)21

21 + 𝑐 because 𝑢 = 𝑥 + 1

Example:

Find ∫_𝑥2𝑥4_+3𝑥3+3𝑥2₊₇ 𝑑𝑥

Let 𝑢 = 𝑥4 + 3𝑥2 + 7

𝑑𝑢

𝑑𝑥 = 4𝑥3+ 6𝑥 𝑑𝑢 = 2(2𝑥3_{+ 3𝑥)𝑑𝑥}

𝑑𝑢

2 = (2𝑥2+ 3𝑥)𝑑𝑥

Replacing all instances of 𝑥 with 𝑢, ⇒ ∫_𝑥42𝑥_+3𝑥3+3𝑥₂₊₇ 𝑑𝑥 =1₂∫_𝑢1𝑑𝑢

1 2∫

1 𝑢𝑑𝑢 =

1

2ln 𝑢 + 𝑐

After the integration, we can replace back all instances of 𝑢 with 𝑥.

⟹ ∫_𝑥2𝑥4_+3𝑥3+3𝑥2₊₇ 𝑑𝑥 =

1 2ln(𝑥

4_{+ 3𝑥}2_{+ 7) + 𝑐}

Example:

Let 𝑢 = 𝑥2

𝑑𝑢 𝑑𝑥 = 2𝑥 𝑑𝑢 = 2𝑥𝑑𝑥

∴ ∫ 𝑒𝑢_{𝑑𝑢 = 𝑒}𝑢_{+ 𝑐}

⇒ 𝑒𝑥2 + 𝑐

Example:

Find ∫(𝑥2+ 1)𝑒(𝑥3_+3𝑥)

𝑑𝑥

Let 𝑢 = 𝑥3 + 3𝑥

𝑑𝑢

𝑑𝑥 = 3𝑥2+ 3 𝑑𝑢 = 3(𝑥2_{+ 1)𝑑𝑥}

𝑑𝑢

3 = (𝑥2+ 1)𝑑𝑥 1

3∫ 𝑒𝑢𝑑𝑢 = 1

3𝑒𝑢+ 𝑐 ⇒1₃𝑒(𝑥3+3𝑥)_{+ 𝑐}

Other Examples of Integration

Find ∫(2𝑥 − 3)2(𝑥 + 1)2_𝑑𝑥

For a problem like this where the power is of small number, it is advisable to expand the function. In this instance, (2𝑥 − 3)2(𝑥 + 1)2 = 4𝑥4 − 4𝑥3− 11𝑥2+ 16𝑥 + 9. Then,

∫(4𝑥4 _{− 4𝑥}3_{− 11𝑥}2_{+ 16𝑥 + 9)𝑑𝑥 =}4

5𝑥5− 𝑥4− 11

The Definite Integral

Given a function 𝑦 = 𝑓(𝑥) (e.g. say, 𝑦 is a cost function and 𝑥 is quantity), the area under the curve between the interval [𝑎 𝑏] is given as:

∫ 𝑓(𝑥)𝑑𝑥_𝑎𝑏 = [𝐹(𝑥) + 𝑐]𝑎𝑏 = 𝐹(𝑏) − 𝐹(𝑎)

Illustration

Find ∫ (𝑥 + 2)𝑑𝑥₁2 We know that

∫(𝑥 + 2)𝑑𝑥 = [1

2𝑥2 + 2𝑥] + 𝑐

Thus, ∫ (𝑥 + 2)𝑑𝑥₁2 = [0.5 ∗ 22_{+ 2 ∗ 2] − [0.5 ∗ 1}1_{+ 2 ∗ 1] = 3.5}

Example:

Find ∫ (𝑥₀1 2 + 𝑥)𝑑𝑥

∫(𝑥2_{+ 𝑥)𝑑𝑥 =}1

3𝑥3+ 1

2𝑥2+ 𝑐

Thus, ∫ (𝑥₀1 2+ 𝑥)𝑑𝑥 = [1₃∗ 13₊1 2∗ 1

2_{] − [0] =}1 3+

1 2=

5 6

Practical example

A manufacturer’s marginal cost function is 𝑑𝑐_𝑑𝑞= 0.6𝑞 + 𝑐. If production is currently set at 𝑞 =

Solution

𝐶 = ∫100(0.6𝑞 + 2)𝑑𝑞

80

𝐶 = [0.3𝑞2_{+ 2𝑞 + 𝑐]} 80 100

= [0.3(100)2+ 2(100)] − [0.3(80)2+ 2(80)]

𝐶 = 1120

∴ The manufacturer would need an additional cost of 1120 to increase quantity from 80 units to 100units.

APPROXIMATE INTEGRATION

Sometimes, it is difficult to integrate a function using the techniques we’ve learnt so far. For example, it will take a considerable effort to find ∫ 𝑒₀2 𝑥2𝑑𝑥. Below is the curve 𝑒𝑥2 between the interval 𝑥 = 0 and 𝑥 = 2.

The area under the curve between this interval is equivalent to the integral ∫ 𝑒₀2 𝑥2𝑑𝑥.

Therefore, if it’s too difficult to integrate ∫ 𝑒₀2 𝑥2𝑑𝑥, we could take a short-cut by rather finding the area under the curve.

The problem we face using this approach is that the shape under the curve might not be a nice one with an already established formula for finding the area, like say, a triangle, rectangle, sphere, etc. In an attempt to still find the area, some scholars proposed dividing the area into sub areas with some triangle, some rectangle etc.

0 10 20 30 40 50 60

0 0.5 1 1.5 2 2.5

y

We will look at one approach to finding the area under a curve using a trapezoid.

TRAPEZIUM RULE

The area of a trapezium is given as: (A) = 1₂ℎ(𝑓1 + 𝑓2)

Given a curve as shown below, we could divide the area under the curve into a number of trapezoids of equal base, find the area of each trapezoid, and sum them together to get an approximate area under the curve. Note the emphasis on ‘approximate’. There will inevitably be some left over areas that won’t be covered by the trapezoids.

Area = 1₂ℎ[(𝑓(𝑎) + 𝑓(𝑎 + ℎ)) + (𝑓(𝑎 + ℎ) + 𝑓(𝑎 + 2ℎ)) + (𝑓(𝑎 + 2ℎ) + 𝑓(𝑎 + 3ℎ)) +

(𝑓(𝑎 + 3ℎ) + 𝑓(𝑎 + 4ℎ) + ⋯ + (𝑓(𝑎 + (𝑛 − 2)ℎ) + 𝑓(𝑎 + (𝑛 − 1)ℎ)) + 𝑓(𝑎 + (𝑛 − 1)ℎ) + 𝑓(𝑏)]

Area = 1₂ℎ[𝑓(𝑎) + 2𝑓(𝑎 + ℎ) + 2𝑓(𝑎 + 2ℎ) + 2𝑓(𝑎 + 3ℎ) + ⋯ + 2𝑓(𝑎 + (𝑛 − 1)ℎ) + 𝑓(𝑏)]

Area = 1₂ℎ[𝑓(𝑎) + 2[𝑓(𝑎 + ℎ) + 𝑓(𝑎 + 2ℎ) + 𝑓(𝑎 + 3ℎ) + ⋯ + 𝑓(𝑎 + (𝑛 − 1)ℎ)] + 𝑓(𝑏)]

Example:

Find 𝑦 = ∫₀1_1+𝑥1 ₂𝑑𝑥 Solution

The actual answer is 0.785, however, this integral is not that easy to find. We could approximate the area under the curve using the trapezoid rule.

The function _1+𝑥1₂ from 𝑥 = 0 to 𝑥 = 1 is as shown below:

0.4 0.5 0.6 0.7 0.8 0.9 1 1.1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

y

Finding the area under the curve using 10 trapezoids of equal base ℎ

ℎ =1 − 0 10 = 0.1

Sum = 15.68

∴ 𝐴𝑟𝑒𝑎 =0.1

2 (15.68) 𝐴𝑟𝑒𝑎 = 0.784

Example: Try the trapezoid rule for ∫ 𝑒₀2 𝑥2𝑑𝑥

𝑥 𝑦 = 1

1 + 𝑥2 𝑘𝑖 ∗

1 1 + 𝑥2

0 1 1

0.1 0.99 2*0.99 =1.98

0.2 0.96 2*0.96 = 1.92

0.3 0.91 2*0.91 = 1.82

0.4 0.86 2*0.86 = 1.72

0.5 0.8 2*0.8 = 1.6

0.6 0.74 2*0.74 = 1.48

0.7 0.67 2*0.67 = 1.34

0.8 0.61 2*0.61 = 1.22

0.9 0.55 2*0.55 = 1.1

SIMPSON’S RULE (Using the parabola)

We can also approximate the area under a curve using a parabola instead of a trapezoid. This is called the Simpson’s rule. The sum of the areas of all parabolas under the Simpson’s rule is given as:

Area = ℎ₃[𝑓(𝑎) + 4𝑓(𝑎 + ℎ) + 2𝑓(𝑎 + 2ℎ) + 4𝑓(𝑎 + 3ℎ) + 2𝑓(𝑎 + 4ℎ) + ⋯ + 2𝑓(𝑎 +

(𝑛 − 2)ℎ) + 4𝑓(𝑎 + (𝑛 − 1)ℎ) + 𝑓(𝑏)

𝑤ℎ𝑒𝑟𝑒; ℎ =𝑏 − 𝑎 𝑛 𝑛 must be𝒆𝒗𝒆𝒏

Example: REVENUE

Use the Simpson’s rule to approximate the total revenue received from the production of 80 units of a product if the values of the marginal revenue function 𝑑𝑟_𝑑𝑞are as follows:

Q units 0 10 20 30 40 50 60 70 80

𝑑𝑟

𝑑𝑞 10 9 8.5 8 8.5 7.5 7 6.5 7

Solution:

ℎ =80 − 0

8 = 10

Ai 1 × 10 4× 9 2× 8.5 4× 8 2× 8.5 4× 7.5 2× 7 4× 6.5 1×7

10 36 17 32 17 30 14 26 7

Ai= 189 Area = ℎ₃× 𝐴𝑖

Area = 10₃ × 189 Area = 630

∴ total revenue received from the production is 630

Example: Use the Simpson’s rule to find ∫ 𝑒₀2 𝑥2𝑑𝑥 and ∫ _1+𝑥1 2𝑑𝑥

Producer and Consumer Surpluses

For any 𝑑𝑞 quantity below 𝑞0, there is a consumer surplus of (𝑝1− 𝑝0)𝑑𝑞 For any 𝑑𝑞 quantity below 𝑞0, there is a producer surplus of (𝑝0− 𝑝2)𝑑𝑞 The total consumer and producer surpluses are given as:

Consumer Surplus = ∫ (𝑝₀𝑞0 1− 𝑝0)𝑑𝑞 Producer Surplus = ∫ (𝑝₀𝑞0 0− 𝑝2)𝑑𝑞

Question:

The demand function for a product is 𝑝 = 100 − 0.05𝑞, where 𝑝 is the price per unit for quantity 𝑞. The supply function is 𝑝 = 10 + 0.1𝑞. Determine the consumer and producer surplus under the market equilibrium.

At equilibrium, 100 − 0.05𝑞 = 10 + 0.1𝑞

0.15𝑞 = 90 𝑞₀ = 600 and 𝑝0 = 10 + 0.1 ∗ (600) = 70

Consumer Surplus = ∫ (𝑝₀𝑞0 1− 𝑝0)𝑑𝑞= ∫₀600(100 − 0.05𝑞 − 70)𝑑𝑞

∫(100 − 0.05𝑞 − 70)𝑑𝑞 = [30𝑞 − 0.025𝑞2_{] + 𝑐}

∫600(100 − 0.05𝑞 − 70)𝑑𝑞

0

= [30 ∗ 600 − 0.025 ∗ 6002_{] − [0] = 9000}

Producer Surplus = ∫ (𝑝₀𝑞0 0− 𝑝2)𝑑𝑞 = ∫₀600(70 − (10 + 0.1𝑞))𝑑𝑞= ∫₀600(60 − 0.1𝑞)𝑑𝑞

∫(60 − 0.1𝑞)𝑑𝑞 = [60𝑞 − 0.05𝑞2_{] = 𝑐}