1. ## find area integration

find the area between the graph of $\displaystyle y= 6x^2 - 42x + 60$ and the x-axis, between x=1 and x=4.

any ideas? thank you

2. Originally Posted by jvignacio
find the area between the graph of $\displaystyle y= 6x^2 - 42x + 60$ and the x-axis, between x=1 and x=4.

any ideas? thank you
Draw a graph and it should be clear that you need to consider two seperate integrals:

1. $\displaystyle \int_1^2 6x^2 - 42x + 60 \, dx$.

2. $\displaystyle \left| \int_2^4 6x^2 - 42x + 60 \, dx \right|$.

3. Originally Posted by mr fantastic
Draw a graph and it should be clear that you need to consider two seperate integrals:

1. $\displaystyle \int_1^2 6x^2 - 42x + 60 \, dx$.

2. $\displaystyle \left| \int_2^4 6x^2 - 42x + 60 \, dx \right|$.
for the graph should i just plug in some values for x then get points for y? being an absolute intergration does it make a difference?

4. Originally Posted by jvignacio
for the graph should i just plug in some values for x then get points for y? being an absolute integer does it make a difference?
If you're studying integral calculus you should know how to draw the graph of a parabola. Especially one whose equation factorises so easily.

I don't know what you mean by absolute integer?
means take the magnitude of the integral (since the integral itself is negative).

5. Originally Posted by mr fantastic
If you're studying integral calculus you should know how to draw the graph of a parabola. Especially one whose equation factorises so easily.

I don't know what you mean by absolute integer?
means take the magnitude of the integral (since the integral itself is negative).
$\displaystyle \int_1^2 6x^2 - 42x + 60 dx$ + $\displaystyle \int_2^4 6x^2 - 42x + 60 dx$

i got -9 for my final answer?

6. Originally Posted by jvignacio
$\displaystyle \int_1^2 6x^2 - 42x + 60 dx$ + $\displaystyle \int_2^4 6x^2 - 42x + 60 dx$

i got -9 for my final answer?
How can an area between the curve and the x-axis be negative?

The answer has to be positive. Did you draw the graph?

It looks like you haven't taken the magnitude of $\displaystyle \int_2^4 6x^2 - 42x + 60 \, dx$ like I said to. But without seeing th details of your calculation it's impossible to know for sure what mistakes you have made.

7. Originally Posted by mr fantastic
How can an area between the curve and the x-axis be negative?

The answer has to be positive. Did you draw the graph?

It looks like you haven't taken the magnitude of $\displaystyle \int_2^4 6x^2 - 42x + 60 \, dx$ like I said to. But without seeing th details of your calculation it's impossible to know for sure what mistakes you have made.
when i integrated $\displaystyle \int_2^4 6x^2 - 42x + 60 \, dx$ that gave me a -20.. so hence my result being a negative but when u say magnitude, what do this mean?

do i need to change the all the signs to its opposite .. - = + and + = - ?

that gives me 31

8. Originally Posted by jvignacio
when i integrated $\displaystyle \int_2^4 6x^2 - 42x + 60 \, dx$ that gave me a -20.. so hence my result being a negative but when u say magnitude, what do this mean?

do i need to change the all the signs to its opposite .. - = + and + = - ?

that gives me 31
The magnitude of -20 is 20.

So yes, 31 is the answer.