Unizor - Definite Integrals - Examples on Newton-Leibniz Formula

Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Examples of Integration Using
Newton-Leibniz Formula

In this lecture we will consider exactly the same examples we used to demonstrate how definite integral can be calculated using its definition as a limit of sums.

Example 1

Find "area under curve" for f(x) = 10x on segment [a=0, b=4].

Solution

The indefinite integral (antiderivative) of function f(x) = 10x is 5x² (we can omit addition of a constant).
This function should be evaluated at the end points of integration resulting in the following:
∫₀⁴10x dx =
= 5·4² − 5·0² = 80−0 = 80
This corresponds to the answer we have obtained previously using limit of Riemann sums.
The end.
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Example 2

Find "area under curve" for f(x) = −x²+1 on segment [a=−1, b=1].

Solution

The indefinite integral (antiderivative) of function f(x) = −x²+1 is −x³/3+x (we can omit addition of a constant).
This function should be evaluated at the end points of integration resulting in the following:
∫₋₁¹(−x²+1) dx =
= [−(1)³/3+(1)] − [−(−1)³/3+(−1)] =
= 2/3 − (−2/3) = 4/3
This corresponds to the answer we have obtained previously using limit of Riemann sums.
The end.
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Example 3

The driver slows its car down by pressing the brakes from initial speed 20 meters per second to complete stop in 10 seconds, reducing its speed by the same value each second (linear dependency of the speed on time).
Find the distance the car covered during this breaking process.

Solution

First of all, we have to find the formula that represents the speed as a function of time.
Since every second the car slows down by the same number of meters per second, and it took 10 seconds to reduce the speed from 20 to 0, the function describing the speed is V(t)=20−2t.
The indefinite integral (antiderivative) of function V(t)=20−2t is 20t−t² (we can omit addition of a constant).
This function should be evaluated at the end points of integration [0,10] resulting in the following:
∫₀¹⁰(20−2t) dx =
= [20·10−10²] − [10·0−0²] =
= 100 − 0 = 100
The distance cover by car during breaking is 100 meters.
The end.
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Example 4

The Hooke's Law tells that a force needed to expand a string from its neutral position linearly depends on the length we expand it: F=K·x, where F is the force, x - the length of expansion and K - coefficient that depends on the physical properties of a spring.
Given a spring with K=0.5 (in newtons per meter).
Determine work W required to expand this spring by 0.1 meter.

Solution

Physical concept of work is defined as a product of a force by a distance this force is applied if this force is constant. In our case the force is changing with distance. To overcome this, we will approach this problem similarly to calculating an area under curve, where curve represents the force.
Partition the distance (spring's expansion) into small intervals and assume that the force is constant on each interval. If the force is a function of distance F(x) then the amount of its work from distance x=0 to distance x=d can be represented as
∫₀^dF(x) dx

Applying this to our case, we have to calculate
∫₀^0.10.5·x dx
The indefinite integral of function 0.5·x is 0.25·x².
Therefore,
W = ∫₀^0.10.5·x dx =
= 0.25·0.1² − 0.25·0² = 0.0025
The work equals to 0.0025(joules)
The end.