Finding the area beneath a curve help

[dc2209]

1. Hi, I have been asked to calculate the area beneath a curve. The curve is
y = 2x^2 + 7x + 24 and the values of x to find the area within are 5 and 2.


3. My attempt stands as follows: A = 5,2∫ (2x^2+7x+24)dx
[ 2x^3/2 + 7x^2/2 + 24x ]
From there I am unsure on what to do as i am unsure on calculus as a whole.



Thanks

 

[HallsofIvy]

1. Hi, I have been asked to calculate the area beneath a curve. The curve is
y = 2x^2 + 7x + 24 and the values of x to find the area within are 5 and 2.


3. My attempt stands as follows: A = 5,2∫ (2x^2+7x+24)dx
[ 2x^3/2 + 7x^2/2 + 24x ]
From there I am unsure on what to do as i am unsure on calculus as a whole.



Thanks

I presume that the "5,2" in front of the integral indicates that you are integrating "from x= 2 to x= 5". Do you know what that means? Yes, 2x^3/3+ 7x^2/2+ 24x+ C (C can be any constant) is the general anti-derivative. Now, evaluate that between x= 2 and x= 5. That is find the value of 2x^3/3+ 7x^2/2+ 24x+ C at x= 5, at x= 2 and subtract the first from the second.

If F'(x)= f then \int f(x)dx= F(x)+ C and \int_a^b f(x)dx= F(b)- F(a)

 

[Student100]

Look at the way you calculated the integral.

Also when calculating a definite integral what makes it different than an indefinite one? Why can you disreguard the arbitary constant?

Yes, 2x^3/3+ 7x^2/2+ 24x+ C (C can be any constant) is the general anti-derivative.

He actually got the indefinite wrong, but yes this is what you should have gotten.

 

[dc2209]

Ok I got (250/2 + 175/2 +24 ) - (16/2 + 28/2 + 24) = 236.5 - 46 = 190.5 units of area?


Does that sound correct?

Thanks,

Ps, would y = -x^2 + 1 use the same method?

 

[Student100]

Ok I got (250/2 + 175/2 +24 ) - (16/2 + 28/2 + 24) = 236.5 - 46 = 190.5 units of area?


Does that sound correct?

Thanks,

Ps, would y = -x^2 + 1 use the same method?

No, look at your orginal calculation, then look at HallsofIvy post. Do you see a difference?

 

[dc2209]

No, look at your orginal calculation, then look at HallsofIvy post. Do you see a difference?

He has misplaced a 3 for a 2, but I didn't use that in my calculation so 190.5 should be correct?

 

[Student100]

He has misplaced a 3 for a 2, but I didn't use that in my calculation so 190.5 should be correct?

His was right, yours was wrong. You raised it to the 3rd power.

 

[dc2209]

His was right, yours was wrong. You raised it to the 3rd power.

Ah I see what you mean.

 

[dc2209]

His was right, yours was wrong. You raised it to the 3rd power.

So I got 151.497 units of area this time using the correct indefinite

 

[Student100]

So I got 151.497 units of area this time using the correct indefinite

Can you post your work? That's incorrect.

Edit: Looking at your previous post you probably forgot the x after the 24 again.

 

[dc2209]

Can you post your work? That's incorrect.

Edit: Looking at your previous post you probably forgot the x after the 24 again.

2x^3/3 + 7x^2/2 + 24x

= [ 250/3 + 175 /2 +24] - [ 16/3 + 28/2 + 24 ]

= 194.83 - 43.33 = 151.497?

 

[Student100]

2x^3/3 + 7x^2/2 + 24x

= [ 250/3 + 175 /2 +24] - [ 16/3 + 28/2 + 24 ]

= 194.83 - 43.33 = 151.497?

24x

Look at your steps very close.

 

[dc2209]

24x

Look at your steps very close.

I'm unsure on calculus, I understand as much that you multiply by the limits of x and then subtract it but with the x I do not know

 

[Student100]

I'm unsure on calculus, I understand as much that you multiply by the limits of x and then subtract it but with the x I do not know

We're past the calculus, you just got to look at your arithmetic, you are adding 24 instead of 24x or 24(5) and 24(2).

 

[dc2209]

We're past the calculus, you just got to look at your arithmetic, you are adding 24 instead of 24x or 24(5) and 24(2).

Sorry I see now.

As a final answer I got 223.499 now, I am really hoping that is correct

 

[Student100]

Sorry I see now.

As a final answer I got 223.499 now, I am really hoping that is correct

That's right, you might want to practice keeping the answers in fraction forum, however. Just keep practicing.