# Something Interesting I Realized About Integrals

• okkvlt
In summary, the speaker discusses a method of finding the integral of a function by using substitution and then finding the limits of integration. They mention that using both values of u1 or both values of u0 as the limits of integration will result in an answer of 0, while using either value of u1 as the upper limit and either value of u0 as the lower limit will result in the correct answer of 20. They also mention that there are a total of 4 ways to correctly get the answer, as long as a value of u1 is used as the upper limit and a value of u0 is used as the lower limit.
okkvlt
suppose i want to find the following integral:
7
$$\int$$x dx
3

now suppose for some demented reason i decided not to do it straightforward and get (49-9)/2=20

instead i use the substitution x=u2+4u+5
giving
u1
$$\int$$(u2+4u+5)(2u+4)du
u0
u1
$$\int$$2u3+12u2+26u+20 du
u0
the indefinite integral is .5u4+4u3+13u2+20u

now its time to find the limits of integration,
u1=(-4$$\pm$$$$\sqrt{24}$$)/2=0.44948974278318,-4.44948974278318

u0=(-4$$\pm$$$$\sqrt{8}$$)/2=-0.5857864376269, -3.4142135623731
what happens when you enter these limits of integration?
Here is what's fascinating:
if i use both values of u1 as the limits of integration, i get 0 (the case is the same with using both values of u0).
if i use either value of u1 as the upper limit and either value of u0 as the lower limit, i get the correct answer of 20. there is a total of 4 ways to correctly get the answer. it doesn't matter which of the values i use as long as i use a value of u1 as the upper limit and a value of u0 as the lower limit.

you can go ahead and test it to see what i mean.

Last edited:

okkvlt said:
suppose i want to find the following integral:
7
$$\int$$x dx
3

now suppose for some demented reason i decided not to do it straightforward and get (49-9)/2=20

instead i use the substitution x=u2+4u+5
giving
u1
$$\int$$(u2+4u+5)(2u+4)du
u0
u1
$$\int$$2u3+12u2+26u+20 du
u0
the indefinite integral is .5u4+4u3+13u2+20u

now its time to find the limits of integration,
u1=(-4$$\pm$$$$\sqrt{24}$$)/2=0.44948974278318,-4.44948974278318

u0=(-4$$\pm$$$$\sqrt{8}$$)/2=-0.5857864376269, -3.4142135623731

what happens when you enter these limits of integration?
Here is what's fascinating:
if i use both values of u1 as the limits of integration, i get 0 (the case is the same with using both values of u0).
Yes, of course. The two values of u1 both correspond to x= 7 and the integral of any function from "7" to "7" is 0.

if i use either value of u1 as the upper limit and either value of u0 as the lower limit, i get the correct answer of 20. there is a total of 4 ways to correctly get the answer. it doesn't matter which of the values i use as long as i use a value of u1 as the upper limit and a value of u0 as the lower limit.
Somewhere in your calculation you are doing the equivalent putting your values of u0 and u1 into the formula u2+ 4u+ 5 and getting values of 7 for both values of u1 and 3 for both values of u0. This behavior is eactly what you should expect.

you can go ahead and test it to see what i mean.

## 1. What are integrals and how are they used in science?

Integrals are mathematical tools used to calculate the area under a curve on a graph. In science, they are used to find the total amount of a quantity over a given period of time or space.

## 2. How are integrals related to derivatives?

Integrals and derivatives are inverse operations of each other. Integrals can be thought of as the reverse of derivatives, where derivatives calculate the rate of change of a quantity, and integrals calculate the total amount of the quantity.

## 3. Can integrals be solved without calculus?

No, integrals are fundamental concepts in calculus and cannot be solved without using calculus techniques.

## 4. Are there different types of integrals?

Yes, there are definite integrals and indefinite integrals. Definite integrals have specific limits of integration and provide a numerical value, while indefinite integrals have no limits and provide a general expression.

## 5. How are integrals used in real-world applications?

Integrals have numerous applications in science, engineering, economics, and other fields. Some examples include calculating the area under a velocity curve to determine displacement, finding the volume of irregularly shaped objects, and determining the total revenue or cost of a business over a period of time.

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