Applying integral calculusSolving differential equations

The area above and below the x axis and the area between two curves is found by integrating, then evaluating from the limits of integration. Integration is also used to solve differential equations.

Part ofMathsCalculus skills

Solving differential equations

We mentioned before about the \(+ c\) term. We are now going to look at how to find the value of \(c\) when additional information is given in the question.

Example 1 (extension)

Find the equation of the curve for which \(\frac{{dy}}{{dx}} = 4{x^3} + 6{x^2}\) and passes through the point \((1,3)\).

Solution

\(y = \int {(4{x^3}} + 6{x^2})dx = {x^4} + 2{x^3} + c\)

Substituting \(x = 1\) and \(y = 3\) (from the coordinate point given in the question):

\(y = {x^4} + 2{x^3} + c\)

\(3 = {1^4} + 2{(1)^3} + c\)

\(3 = 3 + c\)

\(c = 0\)

Therefore the equation of the curve is \(y = {x^4} + 2{x^3}\)

Example 2 (extension)

Find the equation of the curve for which \(\frac{{dy}}{{dx}} = 2x + 1\) and passes through the point \((2,9)\).

Solution

\(y = \int {(2x + 1)} \,dx = {x^2} + x + c\)

Substituting \(x = 2\) and \(y = 9\):

\(9 = 2^{2} + 2 + c\)

\(9 = 4 + 2 + c\)

\(c = 3\)

Therefore the equation of the curve is \(y = {x^2} + x + 3\)

Example 3 (extension)

The gradient of a tangent to a curve is given as \(f\textquotesingle(x)= 6x - \frac{5}{{{x^2}}}\). Find the equation of the curve if it passes through the point \((1,6)\).

Solution

\(f(x) = \int {6x - \frac{5}{{{x^2}}}\,} dx\)

\(= \int {(6x} - 5{x^{ - 2}})dx\)

\(f(x) = \frac{{6{x^2}}}{2} - \frac{{5{x^{ - 1}}}}{{ - 1}} + c\)

\(f(x) = 3{x^2} + \frac{5}{x} + c\)

Substitute when \(x = 1\) and \(y = 6\):

\(6 = 3{(1)^2} + \frac{5}{1} + c\)

\(6 = 8 + c\)

\(c = - 2\)

Therefore the equation of the curve is \(f(x) = 3{x^2} + \frac{5}{x} - 2\)