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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 $$x\geq0$$)

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

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

$$=2+\dfrac {1}{2}$$

$$=\dfrac {5}{2}$$

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 $$v(t)=s'(t)$$

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

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

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 $$x=b$$

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 $$(x_f)$$.

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

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 = $$=A_1+A_2+A_3$$

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

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

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

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