# 991es calculate wrong integrals? int[0,5] |6x^2 - 33x + 45| dx

#### hassan67

##### New member
When solving the following integral: |6x²-33x+45| from 0 to 5
It gives 62.5 on fx-991es plus 'which is wrong and for some reason it gives the same result (62.5) on the fx-991 arx'
But on the 991es application on phone it gives 62.75!

I tried changing the equation a bit and every time it gives the same answer for both but the mentioned equation always give different results

The absolute symbol [imath]\displaystyle | \cdot \cdot \cdot \cdot \cdot|[/imath] looks cute, however it is the chaos of [imath]\displaystyle 99[/imath]% of students' mistakes. I myself still solve many problems falsely because of it.

We have:
[imath]\displaystyle f(x) = \left|6x^2-33x + 45\right|[/imath]

When I was in the Duck class in the kindergarten, I have learnt that [imath]\displaystyle g(x) = |x| = \pm x [/imath] has two signs. Therefore, to avoid making mistakes, find when the function [imath]\displaystyle f(x) [/imath] has a positive sign and when it has a negative sign.

Let [imath]\displaystyle h(x) = 6x^2-33x + 45[/imath].

I will leave it for you as an exercise to show that:

[imath]\displaystyle 6x^2-33x + 45 = (2x - 5)(3x - 9)[/imath]

We want to find the values of [imath]\displaystyle x[/imath] when [imath]\displaystyle (2x - 5)(3x - 9) = 0[/imath].

[imath]\displaystyle x = 2.5, \ \ x = 3[/imath]

I will also leave it for you as an exercise to show that the function [imath]\displaystyle h(x)[/imath] is negative between [imath]\displaystyle x = 2.5 \ \text{and} \ x = 3[/imath], and positive everywhere else.

This information will give us the piecewise form of [imath]\displaystyle f(x)[/imath].

[imath]\displaystyle f(x) =\begin{cases} \ \ \ (6x^2-33x + 45), & \ \text{everywhere else} \ \\-(6x^2-33x + 45), & \ \ 2.5 < x < 3\end{cases}[/imath]

Now we can safely set correct integrals:

[imath]\displaystyle\int_{0}^{5} f(x) \ dx = \int_{0}^{5} \left|6x^2-33x + 45\right| \ dx[/imath]

[imath]= \displaystyle \int_{0}^{2.5} \left(6x^2-33x + 45\right) \ dx + \int_{2.5}^{3} -\left(6x^2-33x + 45\right) \ dx + \int_{3}^{5} \left(6x^2-33x + 45\right) \ dx = 62.75[/imath]

The wrong approach was:

[imath]\displaystyle\int_{0}^{5} \left(6x^2-33x + 45\right) \ dx = 62.5[/imath]

When solving the following integral: |6x²-33x+45| from 0 to 5
It gives 62.5 on fx-991es plus 'which is wrong and for some reason it gives the same result (62.5) on the fx-991 arx'
But on the 991es application on phone it gives 62.75!

I tried changing the equation a bit and every time it gives the same answer for both but the mentioned equation always give different results
The first step to calculate indefinite integrals is to use a "graphic calculator" or wolfram - and plot the function f(x) within the domain of integration. you need to first realize that the domain(of integration) is split into 3 parts :

(0 <= x =< 2.5) & (2.5 <= x =< 3) & (3 <= x =< 5)

Realize that f(x) = 0 or DNE in the middle domain (due to the absolute value sign) - thus value of the integral will be "forced to" zero in this domain (2.5 <= x =< 3).
Carry out the integration on the other two domains and add those up.

Last edited:
When solving the following integral: |6x²-33x+45| from 0 to 5
It gives 62.5 on fx-991es plus 'which is wrong and for some reason it gives the same result (62.5) on the fx-991 arx'
But on the 991es application on phone it gives 62.75!

I tried changing the equation a bit and every time it gives the same answer for both but the mentioned equation always give different results
Please show exactly what you entered on each calculator. It appears that you failed to include the absolute value, or entered it in an inappropriate way.

Realize that f(x) = 0 or DNE in the middle domain (due to the absolute value sign) - thus value of the integral will be "forced to" zero in this domain (2.5 <= x =< 3).
Carry out the integration on the other two domains and add those up.
Professor Khan, I don't agree with you in this part above.

Let [imath]\displaystyle f(x) = \left|6x^2 - 33x + 45\right|[/imath]

And

Let [imath]\displaystyle h(x) = 6x^2 - 33x + 45[/imath]

Since the function [imath]\displaystyle h(x)[/imath] has a negative area between the x-axis and the curve in the domain [imath]\displaystyle 2.5 \leq x \leq 3[/imath], [imath]\displaystyle f(x)[/imath] must have an equal but positive area in the same domain.

Professor Khan, I don't agree with you in this part above.

Let [imath]\displaystyle f(x) = \left|6x^2 - 33x + 45\right|[/imath]

And

Let [imath]\displaystyle h(x) = 6x^2 - 33x + 45[/imath]

Since the function [imath]\displaystyle h(x)[/imath] has a negative area between the x-axis and the curve in the domain [imath]\displaystyle 2.5 \leq x \leq 3[/imath], [imath]\displaystyle f(x)[/imath] must have an equal but positive area in the same domain.
You are correct! Response #3 should be modified as follows:

The first step to calculate indefinite integrals is to use a "graphic calculator" or wolfram - and plot the function f(x) within the domain of integration. you need to first realize that the domain(of integration) is split into 3 parts :

(0 <= x =< 2.5) & (2.5 <= x =< 3) & (3 <= x =< 5)

Realize that f(x) is forced to be positive in the middle domain (due to the absolute value sign) - thus value of the integral will be "forced to" positive in this domain (2.5 <= x =< 3).
Carry out the integration on the three domains. The value of the integral will be calculated negative in the middle domain but needs to be forced to positive. This will be lot clearer if you plot the function f(x) = | 6x^2 - 33x + 45 | with the proper domain.

Thanks Mario.

The first step to calculate indefinite integrals is to use a "graphic calculator" or wolfram - and plot the function f(x) within the domain of integration. you need to first realize that the domain(of integration) is split into 3 parts :

(0 <= x =< 2.5) & (2.5 <= x =< 3) & (3 <= x =< 5)

Realize that f(x) = 0 or DNE in the middle domain (due to the absolute value sign) - thus value of the integral will be "forced to" zero in this domain (2.5 <= x =< 3).
Carry out the integration on the other two domains and add those up.
Here we (or should I say I) go again. I think that students need to think about what is going on instead of just jumping to using software. If a student understands that the their function is a piecewise function and they know how to find the cut-off points then fine, they can use software. In my opinion, this student doesn't understand about the absolute value function being a piecewise function. If the OP did, then they would have input the function correctly w/o absolute value bars and arrived at the same answer regardless of where they checked.
One might state that the OP question is why am I get different answers. If that was the case then their question would have been phrased differently--like why is some of my software not reading the absolute value bar in an integral