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Application Of Definite Integral

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

Definite Integrals Involving Modulus

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 \(x\geq0\))

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

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

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

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

Illustration Questions

The value of  \(I=\int\limits_0^4|x-1|\;dx\) is

A 17

B –24

C 14

D 5

×

To evaluate  \(I=\int\limits_0^4|x-1|\;dx\)  put \(x-1=0\Rightarrow x=1=C\)

 

 \(I=\int\limits_0^1|x-1|\;dx + \int\limits_1^4|x-1|\;dx\)

\(I=\int\limits_0^1-(x-1)\;dx + \int\limits_1^4(x-1)\;dx\)

\(|x-1|=x-1\) if  \(x\in[1, 4]\)

            \(=1-x\) if  \(x\in(0,1)\)

\(=-\dfrac {x^2}{2}+x\Bigg]_0^1+\dfrac {x^2}{2}-x\Bigg]_1^4\)

\(=\left [ \Bigg( \dfrac {-1}{2}+1\Bigg)-0 \right] + \left [ \Bigg( \dfrac {16}{2}-4\Bigg)- \Bigg( \dfrac {1}{2}-1\Bigg) \right]\)

\(=\dfrac {1}{2}+\dfrac {9}{2}\)

\(=5\)

The value of  \(I=\int\limits_0^4|x-1|\;dx\) is

A

17

.

B

–24

C

14

D

5

Option D is Correct

The Net Change Theorem

\(\int\limits_a^bf(x)\;dx=F(b)-F(a)\)  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 \(v(t)=s'(t)\)

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

 

Illustration Questions

The velocity function of a particle is  \(v(t)=3t+7\)  in m/s, find the displacement during time \(2\leq t\leq 5\) (in seconds).

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

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

C \(81\,m\)

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

×

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

\(\Rightarrow s(5)-s(2)=\int\limits_2^5(3t+7)\;dt\)

 

\(=\dfrac {3t^2}{2}+7t\Bigg ]_2^5 \)

\(=\left ( 3×\dfrac {25}{2}+35 \right) - \left ( 3×\dfrac {4}{2}+14 \right) \)

\(=\dfrac {75}{2}+35-6-14\)

\(= \dfrac {75}{2}+15\)

\(= \dfrac {105}{2}\;m\)

The velocity function of a particle is  \(v(t)=3t+7\)  in m/s, find the displacement during time \(2\leq t\leq 5\) (in seconds).

A

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

.

B

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

C

\(81\,m\)

D

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

Option B is Correct

Distance Traveled by a Particle when the Velocity Function is given

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

Displacement  \(=A_1-A_2+A_3\)

distance = \(=A_1+A_2+A_3\)

Illustration Questions

The velocity function of a particle  \(v(t)=4t-1\)  is given in m/s. Find the distance traveled by particle in the time interval \(0\leq t\leq2\).

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

B \(12\,m\)

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

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

×

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

\(=\int\limits_{0}^{2}\,|4t-1\,|\;dt\)

\(=\int\limits_{0}^{1/4}\,|4t-1\,|\;dt + \int\limits_{1/4}^{2}\,|4t-1\,|\;dt\)

\(=\int\limits_{0}^{1/4}\,-(4t-1)\,\;dt + \int\limits_{1/4}^{2}\,(4t-1)\,\;dt\)

\(=\dfrac {-4t^2}{2}+t\Bigg]_0^{1/4}+\dfrac {4t^2}{2}-t\Bigg]_{1/4}^{2}\)

\(=\left ( -2×\dfrac {1}{16}+\dfrac {1}{4} \right)+ \left ( 2×4-2 \right) - \left (2×\dfrac {1}{16}-\dfrac {1}{4} \right)\)

\(=\left (\dfrac {-1}{8}+\dfrac {1}{4}\right)+ \left (6-\left (\dfrac {1}{8}-\dfrac {1}{4}\right) \right)\)

\(=\dfrac {1}{8}+6+\dfrac {1}{8}\)

\(=\dfrac{25}{4}\;m\)

The velocity function of a particle  \(v(t)=4t-1\)  is given in m/s. Find the distance traveled by particle in the time interval \(0\leq t\leq2\).

A

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

.

B

\(12\,m\)

C

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

D

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

Option A is Correct

Acceleration and Velocity

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

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

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

 

Illustration Questions

The acceleration function in  \(m/s^2\)  is given as  \(a(t)=3t+5\) . If the initial velocity  \(v(0)=2\,m/s\) , then find the velocity at time 't'.

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

×

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

Put \(t_1 = 0\) and \(t_2 = t \)

\(\Rightarrow \int\limits_{0}^{t}\,a(t)\;dt=v(t)-v(0)\)

\(=\int\limits_{0}^{t}\,(3t+5)\,dt =v(t)-v(0)\)

\(=\dfrac {3t^2}{2}+5\,t\Bigg]_0^{t}=v(t)-2\, m/s\)

\(\Rightarrow\left (\dfrac {3t^2}{2}+5\,t\right)-0=v(t)-2\, m/s\)

\(\Rightarrow v(t)=\left (\dfrac {3t^2}{2}+5\,t+2\right)m/s\)

The acceleration function in  \(m/s^2\)  is given as  \(a(t)=3t+5\) . If the initial velocity  \(v(0)=2\,m/s\) , then find the velocity at time 't'.

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

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

Option B is Correct

Problems Related to Rate of Change

  • 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 \(x=b\)

Illustration Questions

The linear density of a rod of length 5 m is given by  \(\rho(x)=\left ( 7+\dfrac {1}{\sqrt x} \right)\,kg/m\) where \(x\) is in \('m'\) measured from one end of rod. Find the total mass of the rod.

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

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

C \(351\;kg\)

D \(24\;kg\)

×

\(\int\limits_a^b\,\rho(x)\;dx=m(b)-m(a)\)

\(\Rightarrow \displaystyle\int\limits_0^5\, \left (7+\dfrac {1}{\sqrt {x}} \right)=m(5)-m(0)\)

\(\Rightarrow \, 7x+\dfrac {x^{1/2}}{1/2} \Bigg]_0^5=m(5)-0\rightarrow\) No mass at initial point.

\(\Rightarrow (35+2\sqrt 5)-0=m(5)\)

\(\Rightarrow m(5)=(35+2\sqrt 5)\,kg\)

The linear density of a rod of length 5 m is given by  \(\rho(x)=\left ( 7+\dfrac {1}{\sqrt x} \right)\,kg/m\) where \(x\) is in \('m'\) measured from one end of rod. Find the total mass of the rod.

A

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

.

B

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

C

\(351\;kg\)

D

\(24\;kg\)

Option A is Correct

Work done by Variable Force

Variable Force:

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

Work done by Variable Force:

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

Illustration Questions

The spring force is given as \(F=2x\) where \(x\) is the distance from origin. Calculate the work done by the spring if its length is stretched from \(x_1 = 2\; m\) to \(x_2 = 3 \; m\).

A 5 N

B 6 N

C 7 N

D 8 N

×

The total work done by spring force \(F(x)\) is given as 

\(W_T=\int\limits_{\text {1st element}}^{\text {Last element}}\,F(x)\;dx\)

Since, \(F(x)=2x\) and \(x\) varies from \(x_1 = 2 \; m\) to \(x_2 = 3\; m\) . So work done is given by 

\(W=\displaystyle\int\limits_{2}^{3}\,2x\; dx\)

 \(W = \; 2 \displaystyle\int\limits_2^3 \; x \; dx\)

\(= 2 \left[\dfrac{x^2}{2}\right]_2^3\)

\(= 2 \left[\dfrac{9}{2} - \dfrac{4}{2}\right]\)

\(=2 \left[\dfrac{9}{2} - 2\right]\)

\(= \dfrac{5}{2} × 2 = \; 5 \; N\)

The spring force is given as \(F=2x\) where \(x\) is the distance from origin. Calculate the work done by the spring if its length is stretched from \(x_1 = 2\; m\) to \(x_2 = 3 \; m\).

A

5 N

.

B

6 N

C

7 N

D

8 N

Option A is Correct

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