IBHM 446 472 .pdf
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16 Integration 3 – Applications
1
y3
y xex
3
˛
冮 1e
x
dx
3
y
y xex ex k
3
As was shown earlier, k can be evaluated if the initial condition that fits the
equation is given. In this situation, we could be told that when x 2, y 1
and we can evaluate k.
1
˛
In this situation an explicit
equation in y cannot be
found.
1
1 2e2 e2 k
3
4
1 k 3e2
3
1 k 1.73 p
Therefore the final answer would be
y3
y xex ex 1.74
3
˛
Example
The diagram below shows a tangent to a curve at a point P which cuts the
x-axis at the point A. Given that OA is of length qx, show that the points on the
curve all satisfy the equation
y
y
dy
dx
x qx
P
0
x
qx A
1
a Hence show that the equation is of the form y kx1 q where k is a
˛
constant.
b Given that q is equal to 2, find the equation of the specific curve which
passes through the point (1, 1).
y
The gradient is
¢y
.
¢x
P(x, y)
y
x qx
0 qx
x
(qx, 0)
Therefore the gradient is
Hence
y0
.
x qx
y
dy
dx
x qx
453