# Area under curve problem again

• lionely
In summary, the equation of the curve passing through (-1,0) with a gradient of 3x^2-6x+4 at any point is x^3-3x^2+4x+8. To find the area enclosed by the curve, the x-axis, and the ordinates x=1 and x=2, the equation was integrated from x=1 to x=2, resulting in a value of 2. This value did not match the answer in the book.
lionely

## Homework Statement

Find the equation of the curve which passes through the point (-1,0) and whose gradient at
any point (x,y) is 3x2-6x+4. Find the area enclosed by the curve, the axis of x and the ordinates x=1 and x=2.

. The attempt at a solution

I integrated and got the equation of the curve to be x3 -3x2 +4x +8

Now to find the area I'm not sure what to do, I tried to sketch a graph but I couldn't.

I tried integrating from x=1 to x=2 and I got 2, which isn't the answer in the book :S.

Help greatly appreciated.

lionely said:

## Homework Statement

Find the equation of the curve which passes through the point (-1,0) and whose gradient at
any point (x,y) is 3x2-6x+4. Find the area enclosed by the curve, the axis of x and the ordinates x=1 and x=2.

. The attempt at a solution

I integrated and got the equation of the curve to be x3 -3x2 +4x +8

Now to find the area I'm not sure what to do, I tried to sketch a graph but I couldn't.

I tried integrating from x=1 to x=2 and I got 2, which isn't the answer in the book :S.

Help greatly appreciated.

Which equation did you integrate? You are given the gradient of the curve at any point (x,y). You found the curve. Find the area closed by this curve and the x-axis.

x^3 - 3x^2 +4x +8

lionely said:
x^3 - 3x^2 +4x +8
Show the steps you made when you integrated this from x=1 to x=2 to result in the value 2.

Last edited:
lionely said:
x^3 - 3x^2 +4x +8

Integrate it again from x=1 to x=2. If you integrate 3x^2-6x+4 from x=1 to x=2, then you get the value 2. Knowing this, I asked you which equation did you integrate.

Sorry I intregrated 3x^2-6x+4 for x=1 to x=2 and got 2

## 1. What is the "area under curve problem"?

The "area under curve problem" refers to the mathematical concept of finding the area between a curve and the x-axis on a graph. This problem is commonly encountered in calculus and is used to solve various real-world problems.

## 2. How is the area under a curve calculated?

The area under a curve can be calculated by using integration, which involves finding the antiderivative of the function and evaluating it at the desired boundaries. Alternatively, the area can also be approximated by using numerical methods such as Riemann sums or the trapezoidal rule.

## 3. What are some real-world applications of the area under curve problem?

The area under curve problem has various applications in fields such as physics, economics, and engineering. For example, it can be used to calculate the work done by a force, the profit generated by a company, or the volume of a complex object.

## 4. Can the area under a curve be negative?

Yes, the area under a curve can be negative. This occurs when the function dips below the x-axis, resulting in a negative value for the area. It is important to pay attention to the sign of the area, as it can provide valuable information about the behavior of the function.

## 5. What are some strategies for solving area under curve problems?

Some strategies for solving area under curve problems include identifying the appropriate boundaries, choosing the appropriate method for integration or approximation, and using symmetry or geometry to simplify the problem. It is also important to carefully consider the units of measurement and any given constraints or conditions.

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