Applications of Integration as Anti-Differentiation:

1. Finding function Given  \({dy \over dx}\), the function y=f(x) is found by integrating  \(dy \over dx\)  with respect to x and using the conditions given to calculate the constant of integration.        Step 1: integrate the derivative        Step 2: use the values given to calculate the constant of integration        Step 3: form the function Example Given that  \(dy \over dx\) =2x-3 and that y=12 when x=2, find the equation of y in terms of x.   \(y=\int {dy \over dx}dx \) \( =\int (2x-3)dx \) \( =x^2-3x+C \) \( When \text { }x=2; \)             \( 4-6+C=12 \) \( C=52 \) \( y=x^2-3x+{5 \over 2} \)   Example Given that  \(dy \over dt\) =5t2-2 and that y=10 when t=-1, find the equation of y when t=1.   \( y=\int {dy \over dt}dt \) \( =\int (5t^2-2)dt \) \( ={5 \over 3}t^3-2t+C \) \( When \text { }t=-1; \)             \( -{5 \over 3}+2+C=10 \) \( C={29 \over 3 } \) \( y={5 \over 3}t^3-2t+{29 \over 3} \)     Example 3 Find the equation of a curve with gradient x2-2x and passing through point (1,-1).    \( y=\int (x^2-2x)dx \) \( =13x^3-x^2+C \) \( At (1,-1); \)            \( 13-1+C=-1 \) \( C=-13 \) \( y=13x^3-x^2-13 \)   2. Calculating rate of change Given the rate of change  \(dy \over dt\), the function y=f(t) is found by integrating dydt with respect to t. As above, use the conditions given to calculate the value of the constant of integration.   Example The rate of change of s with respect to t is given by dsdt=8t3 and when t=2, s=2. Find s in terms of t.   \(s=\int {ds \over dt}dt \) \( =8t^3dt \) \( =2t^4+C \) \( When t=2; \)               \( 32+C=2 \) \( C=-30 \) \( s=2t^4-30 \)   3. Other real life examples Example The change in the current of an electric circuit is given by   \({dI \over dt}=-{100t \over t^2} \)   where current, I, is measured in amperes and time t is measured in seconds.  Form the current function I(t) given that the current is 150 amperes at time 2 seconds.   \( {dI \over dt}=-100t^{-2} \) \( I=\int -100t^{-2}dt \) \( =-100t-1-1+C \) \( When t=2; \)           \( 1002+C=150 \) \( C=100 \) \(∴I(t)=100t+100\)   Example A production manager finds that the cost to produce x number of items, in RM per item, is   \({dC \over dx}=3.15+0.004x.\)   Calculate the total cost of producing one hundred items if the fixed costs (that is the cost before production began) is RM450.   \( Cost = \int (3.15+0.004x)dx \) \( =3.15x+0.002x^2+C \) \( When \text { }x=0; C=450 \) \( Then, when \text { } x=100; \) \( Cost =3.15(100)+0.002(100)2+450 \) \( = RM 785 \)

 

Tag Secondary school Average rate of change

Reflection

Given a derivative, how do you find the function using integration?

How do you find a quantity from the given rate of change using integration?

Find the equation of a curve with gradient 3-4x and passing through point (-2,2).

Find the equation of a curve with gradient  \(7+2x+6x^2\)  and passing through point (1,11). 

Find the equation of a curve with gradient \( 2x^2+7x\)  and passing through the origin.

The rate of change of A with respect to t is given by \(dA \over dt\)  =5t4 and when t=-1, A=2. Calculate A when t=-2.

The rate of change of V with respect to t is given by  \({dV \over dt}=t({1 \over t}+{1 \over t^3})\) and when t=1, V=2. Find V in terms of t.