Applications of Integration as Anti-Differentiation:

Applications of Integration as Anti-Differentiation:

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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?

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

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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.

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