Question

The graph in Figure P7.30 specifies a functional relationship between the two variables u and v. (a) Find${\int }_a^{b}udv$.(b) Find ${\int }_b^{a}udv$(c) Find${\int }_a^{b}vdu$.

Short answer

(a) The value of integral is 0.60 J.

(b) The value of integral is -0.60 J.

(c) The value of integral is 1.50 J.

Step-by-step solution

Identification of given data

The coordinates of point a in graph is$(u_a,v_a)=(5cm,-2N)$

The coordinates of point b in graph is$(u_b,v_b)=(25cm,8N)$

The work done by particle is effect of the force on particle to change the position of the particle.

Determination of equation of line

The slope of the line is given as:

$$\begin{gathered} m=\frac{u_b-u_a}{v_b-v_a} \\ m=\frac{8N-(-2N)}{25cm-5cm} \\ m=0.5N/cm \end{gathered}$$

The equation of line is given as:

$u_a=m{v}_a+b.......\left(1\right)$
Substitute all the values in the above equation.

role="math" localid="1663673294539" $$\begin{gathered} -2N=(0.5N/cm)(5cm)+b \\ b=-4.5N \end{gathered}$$

The equation of line by equation (1) is:

$$\begin{gathered} u=mv+b \\ u=(0.5N/cm)v+(-4.5N) \\ u=0.5v-4.5.......\left(2\right) \end{gathered}$$

Determination of integral of line

(a)

The integral is calculated as:

$$\begin{gathered} {\int }_a^{b}udv={\int }_5^{25}(0.5v-4.5)dv \\ {\int }_a^{b}udv={\left(\frac{0.5v^2}{2}-4.5v\right)}_5^{25} \\ {\int }_a^{b}udv=\left(\frac{0.5{\left(25\right)}^2}{2}-4.5\left(25\right)\right)-\left(\frac{0.5{\left(5\right)}^2}{2}-4.5\left(5\right)\right) \\ {\int }_a^{b}udv=60N.cm \\ {\int }_a^{b}udv=(60N.cm)\left(\frac{1m}{100cm}\right) \\ {\int }_a^{b}udv=0.60N.m \\ {\int }_a^{b}udv=0.60J \end{gathered}$$

Therefore, the value of integral is$0.60J.$

Determination of integral of line 

(b)

The value of integral is calculated as:

$$\begin{gathered} {\int }_b^{a}udv=-{\int }_a^{b}udv \\ {\int }_b^{a}udv=-(0.60J) \\ {\int }_b^{a}udv=-0.60J \end{gathered}$$

Therefore, the value of integral is$-0.60J.$

Determination of integral of line 

(c)

The equation in terms of u is given as:

$$\begin{gathered} u=0.5v-4.5 \\ v=2u+9 \end{gathered}$$

The integral is calculated as:

localid="1663676187862" $$\begin{gathered} {\int }_a^{b}vdu={\int }_{-2}^8(2u+9)du \\ {\int }_a^{b}vdu={(u^2+9u)}_{-2}^8{\int }_a^{b}vdu=\left({\left(8\right)}^2+9\left(8\right)\right)-\left({(-2)}^2+9(-2)\right) \\ {\int }_a^{b}vdu=150N.cm \\ {\int }_a^{b}vdu=(150N.cm)\left(\frac{1m}{100cm}\right) \\ {\int }_a^{b}vdu=1.50N.m \\ {\int }_a^{b}vdu=1.50J \end{gathered}$$

Therefore, the value of integral is$1.50J.$