Learn definite integral of absolute value & velocity and acceleration calculus, Practice net change theorem & rate of change problems in calculus.

To evaluate

\(I=\displaystyle\int_a^b|f(x)|\;dx\) we split the integral by solving \(f(x)=0\) and if the solution is in the interval [a, b] then we write the integral as

\(I=\displaystyle\int_a^c|f(x)|\;dx+ \displaystyle\int_c^b|f(x)|\;dx\)

where 'C' is the solution to the equation \(f(x)=0\). If there are more than one values of 'C' we split the integrals accordingly, suppose there are two values of C, we write

\(I=\displaystyle\int_a^{C_1}\;f(x)\;dx+ \displaystyle\int_{C_1}^{C_2}\;f(x)\;dx+ \displaystyle\int_{C_2}^b\;f(x)\;dx\)

\(f(x)\) changes behavior at \(x=C\).

Now the sign of \(f(x)\) will be fixed in \((a,c)\) and \((c,b)\). So accordingly we write \(|f(x)|\) as \(f(x)\) or \(-f(x)\)

e.g.

\(=\displaystyle\int_{-2}^{1}\;|\,x\,|\;dx= \displaystyle\int_{-2}^{0}\;|\,x\,|\;dx+ \displaystyle\int_{0}^{1}\;|x|\;dx\)

\(=\displaystyle\int_{-2}^{0}\;-x\;dx+ \displaystyle\int_{0}^{1}\;\,x\,\;dx\)

(\(|x|=-x\) if \(x<0\)

\(=x\) If \(x\geq0\))

\(=\dfrac {-x^2}{2}\Bigg]_{-2}^{0}+ \dfrac {x^2}{2}\Bigg]_{0}^{1}\)

\(=-\dfrac {1}{2} \left [ 0-(-2)^2 \right]+\left (\dfrac {1}{2}-0\right)\)

\(=2+\dfrac {1}{2}\)

\(=\dfrac {5}{2}\)

\(\int\limits_a^bf(x)\;dx=F(b)-F(a)\) where \(f(x)=F'(x)\)

\(\Rightarrow\int\limits_a^b\;F'(x)\;dx=F(b)-F(a) \)

- \(F'(x)\) is the rate of change of \(F(x)\) with respect to \(x\).
- Integral of rate of change is the net change.
- If an object moves along a straight path with position function \(s(t)\), then velocity \(v(t)=s'(t)\)

\(\therefore \int\limits_{t_1}^{t_2}v(t)\;dt=s(t_2)-s(t_1)\)

A \(\dfrac {75}{2}\,m\)

B \(\dfrac {105}{2}\,m\)

C \(81\,m\)

D \(\dfrac {521}{3}\,m\)

- If the mass of a rod measured from left end point to a point \(x\) is \(m(x)\) then the linear density

\(\rho(x)=m'(x)\), so

\(\int\limits_a^b\,\rho(x)\;dx=m(b)-m(a)\) is the mass of rod that lie between \(x=a\) and \(x=b\)

A \(35+2\sqrt 5\;kg\)

B \(28+\sqrt 5\;kg\)

C \(351\;kg\)

D \(24\;kg\)

It is a force whose magnitude is an function of position /displacement, i.e., \(F(x)\).

- Suppose a force is acting on a body which is not constant but depends upon displacement.
- Work done by constant force is expressed as the product of force and displacement, only when force is constant.

\(W=F_{\text {constant}}×\text {Displacement}\)

- Thus, this expression can't be used to calculate work done by variable force.
- Two solve this problem, the displacement is divided into very small parts.
- Since, the parts are so small, so we can assume that force during that part of displacement is constant.
- So, summing up the work done by these small displacements, gives the total work.
- Consider an example, a force \(F=k\,x\) is acting on a body where \(x\) is the distance from origin.

- As we move from A to B, the value of the force changes (increases).
- So, to calculate the work done by force from A to B, we can't take any value of force.
- To calculate total work for this, we divide AB into as many parts as possible.

- Now, considering an element of displacement of thickness \(dx\) at a distance \(x\).
- This element is so small that we can assume force at \(x\)to be \(k\,x\) and force at \((x+dx)\) to be \(k(x+\Delta x)\simeq kx\) as \(\Delta x\to 0\).
- So, work done by small element \(dx\) is

\(W=\vec F \cdot d\vec x\)

- Since the displacement and force are opposite, making an angle of 180°, so

\(W=F \,dx\,×cos\,180°\)

\(W=- F \cdot dx\)

- So, the total work done from all parts of displacement from A to B.

\(W_ T=-\sum \limits_{\text {from 1st element}}^{\text {to last element}}F.dx\)

- This can also be calculated by using calculus.

\(W_ T=-\int\limits_{\text {from 1st element}}^{\text {to last element}}\;F\;dx\)

- In short, we can say that to find total work done by variable force following steps can be followed:

- Write \(F\) as a function of \(x\).
- Write \(x\)as \(dx\).
- Integrate the force from 1st element \((x_i)\) to final element \((x_f)\).

- Distance traveled = \(\int\limits_{t_1}^{t_2}| v(t) |\;dt\)

Displacement \(=A_1-A_2+A_3\)

distance = \(=A_1+A_2+A_3\)

A \(\dfrac {25}{4}\,m\)

B \(12\,m\)

C \(\dfrac {17}{2}\,m\)

D \(\dfrac {92}{5}\,m\)

Acceleration \(=a(t)\) is the rate of change of velocity

\(\Rightarrow a(t)=v'(t)\)

\(\Rightarrow \int\limits_{t_1}^{t_2}\,a(t)\,dt=v(t_2)-v(t_1)\)

A \((4t^2-5t)\,m/s\)

B \(\Big(\dfrac {3t^2}{2}+5t+2\Big)\;m/s\)

C \((2t^2+t+1)\;m/s\)

D \((7t^2-t-11)\;m/s\)