Find the value of this definite integral

[chwala]

Homework Statement:

find the definite integral of
$\int_0^5 -x^3+3x^2+6x-8\,dx$

Relevant Equations:

integration

$\int_0^5 [-x^3+3x^2+6x-8\,]dx$
$\int_0^1 [-x^3+3x^2+6x-8\,]dx= |-\frac {17}{4}|$
$\int_1^4 [ -x^3+3x^2+6x-8\,]dx= 16$
$\int_4^5[-x^3+3x^2+6x-8\,]dx= |-\frac {49}{4}|$
Therefore, total area is $|-\frac {17}{4}|+ 16+|-\frac {49}{4}|=32.5$

now where my problem is,... my colleague indicated the answer as $36.75$

who is right, or where have i made a blunder?

 

[chwala]

 

[chwala]

The above is the solution from a colleague...

 

[fresh_42]

Whatever you have done there, means: I have no idea where you got the numbers from, the result is $3.75.$ If you meant to calculate an area, then the result is different. Which are the zeros of the integrand?

 

[Delta2]

Whatever you have done there, means: I have no idea where you got the numbers from, the result is $3.75.$

Yes i think the problem is not well stated, it first asks for the integral but then i think it means to calculate the area between the graph of the function and the x-axis, so i think the problem wants us to calculate $$\int|-x^3+3x^2+6x-8|dx$$. But I am afraid it isn't that which the OP calculates ...

 

[Delta2]

The above is the solution from a colleague...

I think your colleague is right. The problem (though it is not well stated as i said in post #5) i think wants us to calculate $$\int_0^5 |-x^3+3x^2+6x-8|dx=\int_0^1|f(x)|dx+\int_1^4|f(x)|dx+\int_4^5|f(x)|dx$$ while you calculate something different, i think you calculate $$|\int_0^1f(x)dx|+|\int_1^4 f(x)dx|+|\int_4^5 f(x)dx|$$ $$f(x)=-x^3+3x^2+6x-8$$.
These above two integrals are in general different because as you might already know in general it is $$|\int_a^b f(x)dx|\neq \int_a^b|f(x)|dx$$

 

[chwala]

Yes i think the problem is not well stated, it first asks for the integral but then i think it means to calculate the area between the graph of the function and the x-axis, so i think the problem wants us to calculate $$\int|-x^3+3x^2+6x-8|dx$$. But I am afraid it isn't that which the OP calculates ...

correct. that is the correct solution...i see my blunder...it does not really make sense trying to break them into parts...the function is continuous and the $x-axis$ is a mirror line that will balance out the values, we therefore integrate normally and plug in the limits to get the required value of $3.75$

 

[chwala]

Whatever you have done there, means: I have no idea where you got the numbers from, the result is $3.75.$ If you meant to calculate an area, then the result is different. Which are the zeros of the integrand?

kindly elaborate ...finding the area and finding the definite integrals given limits, is that not one and the same?secondly i would like to know why my solution is wrong, can one not integrate partially and add up the sums...why is my post $1$ wrong...

 

[chwala]

Whatever you have done there, means: I have no idea where you got the numbers from, the result is $3.75.$ If you meant to calculate an area, then the result is different. Which are the zeros of the integrand?

the graph cuts the $x-axis$ at $x=-2, 1$ and $4$ that's why i was thinking of integrating partially...

 

[chwala]

Let me ask this, finding the area of $f(x)$ under the curve from the points $x=0$ and $x=5$, is this not the same as finding the definite integral of $f(x)$ under the same limits?...if they are the same, then my second question is, why is my post $1$ wrong?

 

[Delta2]

Let me ask this, finding the area of $f(x)$ under the curve from the points $x=0$ and $x=5$, is this not the same as finding the definite integral of $f(x)$ under the same limits?...

It is the same thing if the function is strictly positive or strictly negative but if the function is "mixed" i.e has parts where it is negative and other parts where it is positive then it is not the same thing.

In any case the area between the graph of the function and the x-axis is given by the integral $$\int_a^b|f(x)|dx$$.

 

[chwala]

now, assuming that he was correct, which is the correct solution, or did i make a mistake somewhere in my working of post $1$?thanks

 

[chwala]

It is the same thing if the function is strictly positive or strictly negative but if the function is "mixed" i.e has parts where it is negative and other parts where it is positive then it is not the same thing.

In any case the area between the graph of the function and the x-axis is given by the integral $$\int_a^b|f(x)|dx$$.

got it...

 

[fresh_42]

kindly elaborate ...finding the area and finding the definite integrals given limits, is that not one and the same?

Have a read here: https://www.physicsforums.com/threads/double-integrals-do-areas-cancel.995393/

I would a) integrate the formula without boundaries, then b) determine the zeros of the polynomial, c) calculate the three integrals $0\to 1$, $1 \to 4$, $4\to 5$, d) take the absolute values of the three, and finally e) add those numbers.

 

[Delta2]

I would a) integrate the formula without boundaries, then b) determine the zeros of the polynomial, c) calculate the three integrals $0\to 1$, $1 \to 4$, $4\to 5$, d) take the absolute values of the three, and finally e) add those numbers.

Then i think you would do the same mistake as @chwala . You would calculate $$\sum_i|\int_{a_i}^{a_{i+1}}f(x)dx|$$ while the correct is $$\sum_i\int_{a_i}^{a_{i+1}}|f(x)|dx$$ ($a_i$ the roots of the polynomial).

 

[fresh_42]

Then i think you would do the same mistake as @chwala . You would calculate $$\sum_i|\int_{a_i}^{a_{i+1}}f(x)dx|$$ while the correct is $$\sum_i\int_{a_i}^{a_{i+1}}|f(x)|dx$$ ($a_i$ the roots of the polynomial).

That makes no difference, since $[a_i,a_{i+1}]$ doesn't contain zeroes. I even think you are wrong! You changed the function, while I only considered the areas. I do not see any justification to change the integrand!

 

[Delta2]

That makes no difference, since [ai,ai+1] doesn't contain zeroes.

Well, i have to admit you are right here, in the case $a_i,a_{i+1}$ are successive roots of the polynomial, the two expressions i wrote are the same thing.

You changed the function, while I only considered the areas. I do not see any justification to change the integrand!

Isn't the area between the graph of a function and the x-axis given by $$\int_a^b|f(x)|dx$$? That's why i changed the integrand to the absolute value.

 

[Delta2]

ok well @chwala seems you did a mistake in calculation of $$\int_1^4 -x^3+3x^2+6x-8 dx$$ the correct value is $\frac{81}{4}$ and not 16.

 

[chwala]

Thanks delta and fresh_ bingo guys...