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


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