Calculus problem.?

Question

Find the exact area between f(x)=x^3-7x^2+10x, the x-axis, x=0 and x=5.

No. It is not right answer. Using intergal fomula from 0 to 5 cannot find the right answer.

IF THE QUESTION IS:
"Find intergral [o to 5](x^3-7x^2+10x)dx exactly and interpret this intergal in terms of areas", then all of you have the right answer.

The right answer is 253/12. But I don;t know how. If just finding antidirivative (from 0 to 5 ), it's won't be problem.

Answer 1

Find the exact area between f(x) = x³ - 7x² + 10x, the x-axis, x=0 and x=5.

f(x) = x³ - 7x² + 10x
= x(x² - 7x + 10)
= x(x - 2)(x - 5)
= 0 when x = 0, x = 2 and x = 5

So Area = |∫[from x = 0 to 2] x³ - 7x² + 10x dx|
+ |∫[from x = 2 to 5] x³ - 7x² + 10x dx|

= | [¼x^4 - 7/3x³ + 10/2x² ][from x = 0 to 2] |
+ | [¼x^4 - 7/3x³ + 10/2x² ][from x = 2 to 5] |

= | (4 - 56/3 + 20) - (0) | + | (625/4 - 875/3 + 125) - (4 - 56/3 + 20)|

= | 5⅓ | + | -10 5/12 - 5⅓ |

= 5⅓ + 15¾

= 21 1/12

= 253/12

Answer 2

f(x) = x^3 - 7x^2 + 10x

Your first step is to determine whether all of f(x) is above the x-axis. This is done by equation f(x) to 0, and is an important step to determine the bounds of integration.

x^3 - 7x^2 + 10x = 0
x(x^2 - 7x + 10) = 0
x(x - 5)(x - 2) = 0
Therefore, x = {0, 2, 5}

As you can see, the graph crosses the x-axis between our given bounds x = 0 and x = 5, so you're in actuality dealing with two functions:
f(x) = x^3 - 7x^2 + 10x, and
g(x) = 0

From 0 to 2, one of the functions will be higher than the other, and the same thing goes from 2 to 5. To determine which function is greater on the interval 0 to 2, you have to test any number in that interval; let's choose 1.
f(1) = 1 - 7 + 10 = 4, and g(1) = 0. Therefore, f(x) is higher from 0 to 2.
Now, let's test a number between 2 and 5. Let's choose 3.
f(3) = 3^3 - 7(3)^2 + 10(3) = 27 - 63 + 30 = -5, and
g(3) = 0, Therefore, the curve g(x) is higher. Thus, our area problem is calculated as follows.

A = Integral (0 to 2, [f(x) - g(x)] dx) + Integral (2 to 5, [g(x) - f(x)])dx

When calculating the area, it is always "higher curve minus lower curve". Let's calculate this now.

A = Integral (0 to 2, [x^3 - 7x^2 + 10x])dx +
Integral (2 to 5, (0 - [x^3 - 7x^2 + 10x])dx

A = Integral (0 to 2, [x^3 - 7x^2 + 10x])dx +
Integral (2 to 5, -x^3 + 7x^2 - 10x)dx

And we calculate them individually.

A = [(1/4)x^4 - (7/3)x^3 + 5x^2] {evaluated from 0 to 2} +
[(-1/4)x^4 + (7/3)x^3 - 5x^2] {evaluated from 2 to 5}

Which gives us

A = [ { (1/4)2^4 - (7/3)2^3 + 5(2)^2 } - {0 + 0 - 0} ] +
[ { (-1/4)5^4 + (7/3)5^3 - 5(5)^2 } - { (-1/4)2^4 + (7/3)2^3 - 5(2^2) } ]

Which meshes into

A = [4 - 56/3 + 20] + [ { (-625)/4 + 875/3 - 125 } - { -4 + 56/3 - 20 }]

A = 24 - 56/3 + -625/4 + 875/3 - 125 + 24 - 56/3
A = -77 + 763/3 - 625/4

Let's put all of these over a common factor, 12.

A = -924/12 + 3052/12 - 1875/12
A = 2128/12 - 1875/12 = 253/12

The key thing to know about this question is that even though it flat out says the bounds are x = 0 and x = 5, it doesn't necessarily mean you plug those in as your bounds of integration and solve.

Also, the method of solving the integral of the "higher function minus the lower function" method is a great means of avoiding to resort to absolute value to determine negative area.

Answer 3

That's about as easy as an integration problem can get. It's just integral of f(x) evaluated from 0 to 5:

F(x) = 1/4x^4 - 7/3x^3 + 5x^2 + C

F(0) = C
F(5) = 25(1/4(25) - 7/3(5) + 5) + C
= 25(25/4 - 35/3 + 5) + C
= 25(75/12 - 140/12 + 60/12) + C
= 25(-5/12) + C
= -125/12 + C

F(5) - F(0) = -125/12 = -10 5/12

So the curve must be more below the x-axis between 0 and 5, that is, there is more "negative area" than positive area.

If the area below the curve should be counted as positive area as well, then you'd need to figure out where the curve crosses the x-axis by finding the zeros of the function on the interval [0,5] and then integrate piecewise. I don't feel like doing that.

Answer 4

Note that f(x)=x(x-2)(x-5). It is not difficult to see that

- f(x)>0 when 0 - f(x)<0 when 2 - f(x)=0 when x=0, x=2 or x=5.

Then the exact area is:

(Integral of f from 0 to 2)+(integral of -f from 2 to 5)

Answer 5

∫from 0 to 5 x^3-7x^2 +10x dx
= x^4 /4 -7x^3/3 +10x^2 /2 from 0 to 5
=5^4/4 -7(5^3)/3 +5(5^2) .

Answer 6

That's just about one of the easiest problems in integral calculus.

Are you sure that you shouldn't just do it?