Properties of Indefinite Integrals

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Following properties of indefinite integrals arise from the Constant Multiply and Sum rules for derivatives.

Property 1. If `a` is some constant then `int a*f(x)dx=a int f(x)dx +C`. In other words cosntant can be factored out of integral sign.

Property 2. For functions `f(x)` and `g(x)` `int (f(x)+-g(x))dx=int f(x)dx+-int g(x)dx`. Integral of sum (difference) is sum (difference) of integrals. This formula can be easily extended to any number of functions.

Since differentiation and integration are inverse processes then we have additional two properties.

Property 3. `d int f(x)dx=f(x)`. Derivative of integral of a function is function itself.

Property 4. In notation of differentials we have that `dF=F'dx=fdx`, so `int dF(x)=F(x)+C`. Integral of derivative of function is function itself plus some arbitrary constant.

Example 1. Find `F(x)=int f(x)dx` if `f(x)=e^x+x^(-2/3)-5/(1+x^2)`.

Using table of antiderivatives and Properties 1 and 2 we have that `F(x)=int(e^x+x^(-2/3)-5/(1+x^2))dx=int e^xdx+int x^(-2/3)dx-5int 1/(1+x^2)dx=`

`=e^x+(x^(-2/3+1))/(-2/3+1)+5arctan(x)+C`.

So `F(x)=e^x+3x^(1/3)+5arctan(x)+C`.

Example 2. Find `f(x)` if `f'(x)=cos(x)+sin(x)-5x^2` and `f(0)=2`.

According to Property 4 `f(x)=int f'(x)dx+C`.

So `f(x)=int(cos(x)+sin(x)-5x^2)dx=int cos(x)dx+int sin(x)dx-5 int x^2dx=`

`=sin(x)-cos(x)-5/3 x^3+C`.

Since `f(0)=2` then `2=sin(0)-cos(0)-5/3 (0)^3+C` or `C=3`.

Therefore `f(x)=sin(x)-cos(x)-5/3x^3+3`.

Example 3. Find `int (2x^2+1)^3dx`.

Using the formula of Binom of Newton we can rewrite integrand as follows:

`int (2x^2+1)^3dx=int ((2x^2)^3+3(2x^2)^2*1+3*2x^2*1^2+1^3)dx=int (8x^6+12x^4+6x^2+1)dx=`

`=8/7 x^7+12/5 x^5+2x^3+x+C`.

Example 4. Find `int (1+sqrt(x))^4dx`.

Using formula of Binom of Newton we have that

`int (1+sqrt(x))^4dx=int(1+4sqrt(x)+6x+4xsqrt(x)+x^2)dx=`

`=int 1dx+4int x^(1/2)dx+6int xdx+4int x^(3/2)dx+int x^2dx=x+8/3 x^(3/2)+3x^2+8/5 x^(5/2)+1/3 x^3+C`.

Example 5. Find `int ((x+1)(x^2-3))/(3x^2) dx`.

We have that `int (x^3+x^2-3x-3)/(3x^2)dx=int (1/3 x+1/3-1/x-1/x^2)dx=1/3 int xdx+1/3 int 1*dx-int 1/x dx-int 1/x^2 dx=`

`=1/6 x^2+1/3 x-ln|x|+1/x+C`.

Example 6. Find `int ((x-sqrt(x))(1+sqrt(x)))/(root(3)x)dx`.

We have that `int ((x-sqrt(x))(1+sqrt(x)))/(root(3)x)dx=int (x+xsqrt(x)-sqrt(x)-x)/(root(3)x)dx=int (xsqrt(x)-sqrt(x))/(root(3)x)dx=int (x^(3/2)-x^(1/2))/(x^(1/3))dx=`

`=int (x^(7/6)-x^(1/6))dx=int x^(7/6) dx-int x^(1/6)dx=6/13 x^(13/6)-6/7 x^(7/6)+C`.

Example 7. Find `f(x)` if `f''(x)=cos(x)+x^(-3/5)` and `f(0)=2` and `f(1)=3`.

We have that `f'(x)=int f''(x)dx`.

That's why `f'(x)=sin(x)+(x^(-3/5+1))/(-3/5+1)+C` or `f'(x)=sin(x)+5/2 x^(2/5)+C`.

Now `f(x)=int f'(x)dx` , so

`f(x)=-cos(x)+5/2 (x^(2/5+1))/(2/5+1)+Cx+D` or `f(x)=-cos(x)+25/14 x^(7/5)+Cx+D`.

Note that both `C` and `D` are constants.

Since `f(0)=2` then `2=-cos(0)+25/14 (0)^(7/5)+C*0+D` and `D=3`.

Now, `f` has the following form: `f(x)=-cos(x)+25/14 x^(7/5)+Cx+3`.

Since `f(1)=3` then `3=-cos(1)+25/14 (1)^(7/5)+C*1+3` or `C=cos(1)-25/14`.

Therefore `f(x)=-cos(x)+25/14 x^(7/5)+(cos(1)-25/14)x+3`.

Example 8. A particle moves in a straight line and has acceleration given by `a(t)=2t+3`. It is initial velocity is `v(0)=-4 m/s` and its initial dispalcement is `s(0)=10\ cm`. Find its position function.

Since `v'(t)=a(t)=2t+3` then integration gives `v(t)=2 t^2/2+3t+C=t^2+3t+C`.

We have `v(0)=-4`, so `-4=0^2+3*0+C` or `C=-4`, therefore `v(t)=t^2+3t-4`.

Since `s'(t)=v(t)=t^2+3t-4` then integration gives `v(t)=(t^3)/3+3 (t^2)/2-4t+D`.

We have `s(0)=10`, so `10=1/3*0^3+3/2*0^2-4*0+D` or `D=10`.

Therefore position function is `s(t)=1/3 t^3+3/2 t^2-4t+10`.

Example 9. A ball is thrown upward with a speed of `64\ ft / s` from the edge of a cliff 464 ft above the ground. Find its height above the ground `t` seconds later. When does it reach its maximum height? When does it hit the ground?

An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by `g`.

For motion close to the earth we may assume that `g` is constant, its value being about `9.8 m/s^2` (or `32 ft/s^2`).

The motion is vertical and we choose the positive direction to be upward. At time `t` the distance above the ground is `s(t)` and the velocity `v(t)` is decreasing.
Therefore, the acceleration `a(t)` must be negative and we have `a(t)=(dv)/(dt)=-32`.

Taking integral, we have `v(t)=-32t+C`.

From the fact that `v(0)=64` we have that `64=-32*0+C` or `C=64`.

Therefore, `v(t)=-32t+64`.

The maximum height will be reached when `v(t)=0` or `t=64/32=2\ s`.

Since `s'(t)=v(t)`, we integrate again and obtain `s(t)=-16t^2+64t+D`.

Using the fact that `s(0)=464` we have that `464=-16*0^2+64*0+D` or `D=464`.

Therefore, `s(t)=-16t^2+64t+464`.

The expression for `s(t)` is valid until the ball hits the ground. This happens when `s(t)=0`, that is, when `-16t^2+64t+464=0` or `t^2-4t-29=0`.

Using the quadratic formula to solve this equation, we get `t=2+-sqrt(33)`.

We reject the solution with the minus sign since it gives a negative value for time `t`.
Therefore, the ball hits the ground after `2+sqrt(33)~~7.7\ s.`