Recall: When we differentiate a function, we obtain a formula for the gradient of the tangent of the curve.

Therefore, if we know the formula for the gradient, we need to integrate to determine the original function. This integration, due to the constant of integration, will give us a family of functions with the stated gradient. In order to determine the specific function, we would need to use the coordinates of a point which lies on the function.

All of the functions in the diagram have a gradient of \(3x^2-4x\) but only one passes through the point (0, 2)

LESSON 1

The gradient of a particular curve is given by the formula \(3x^2-2x\). Given that this curve passes through the point (2, 5), find the equation of the curve.

SOLUTION

\[y=\int {dy\over dx} dx\]

\[y=\int(3x^2-2x)dx\]

\[y={3x^{2+1}\over 2+1}-{2x^2\over2}+c\]

\[y=x^3-x^2+c\]

- Determine the value of the constant.

Since the curve passes through (2, 5), this point must satisfy the equation of the curve.

\[5=2^3-2^2+c\]

\[1=c\]

The equation of the curve is

\[y=x^3-x^2+1\]

LESSON 2

The curve for which \(\displaystyle {dy\over dx}=-x+a\), where \(a\) is a constant, has a stationary point at (2, 3). Find the equation of the curve.

SOLUTION

(2, 3) is a stationary point therefore this point

\[{dy\over dx}=0\]

\[-x+a=0\]

\[-2+a=0\]

\[a=2\]

Thus, for the curve \(\displaystyle {dy\over dx}=-x+2\)

\[y=\int (-x+2)dx\]

\[y =-{x^{1+1}\over 1+1}+2({x^{0+1}\over 0+1})+c\]

\[y =-{x^2\over2}+2x+c\]

\[y=-{x^2\over2}+2x+c\]

Curve passes through (2, 3) so

\[3=-{2^2\over2}+2(2)+c\]

\[3=-2+4+c\]

\[1=c\]

The equation of the curve is

\[y=-{x^2\over2}+2x+1\]