Additional Problems - Chapter 3 - Integral Calculus I1 - 12th Maths Guide Guide Samacheer Kalvi Solutions

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Additional Problems
I. One Mark Questions
Choose the correct answer.
Question 1.
The area bounded by the curve $y=e^x$, the $x$-axis and the lines $x=0$ and $x=3$ is
(a) $\mathrm{e}^3-1$
(b) $\mathrm{e}^3+1$
(c) $\mathrm{e}^3$
(d) $\mathrm{e}^3-2$
Answer:
(a) $\mathrm{e}^3-1$
Hint:
Area $=\int_0^3 e^x d x=\left(e^x\right)_0^3=e^3-1$
Question 2.
The area bounded by the demand curve $\mathrm{xy}=1$, the $\mathrm{x}$-axis, $\mathrm{x}=1, \mathrm{x}=5$ is
(a) $\log 5$
(b) $\log \frac{1}{5}$
(c) $\log 4$
(d) $\frac{1}{5} \log 2$
Answer:
(a) $\log 5$
Hint:
Area $=\int_1^5 \frac{1}{x} d x=(\log x)_1^5=\log 5$
Question 3 .
If the marginal cost function $\mathrm{MC}=7-4 \mathrm{x}$ then the cost function is
(a) $7 x-2 x^2+k$
(b) $7-4 \mathrm{x}^2$
(c) $\frac{7}{x}-4$
(d) $7 x-4 x^2$
Answer:

(a) $7 x-2 x^2+k$
Hint:
$
\begin{aligned}
& \mathrm{C}=\int \mathrm{MCdx} \\
& =\int(7-4 \mathrm{x}) \mathrm{dx} \\
& =7 \mathrm{x}-2 \mathrm{x}^2+\mathrm{k}
\end{aligned}
$
Question 4.
The marginal revenue $\mathrm{R}^{\prime}(\mathrm{x})=\frac{1}{(x+1)}$ Then the revenue function is
(a) $\frac{-1}{x+1}$
(b) $\frac{1}{(x+1)^2}$
(c) $\log \frac{1}{(x+1)}$
(d) $\log |\mathrm{x}+1|+\mathrm{k}$
Answer:
(d) $\log |\mathrm{x}+1|+\mathrm{k}$
Hint:
Revenue function $\mathrm{R}=\int \frac{1}{x+1} d x=\log |\mathrm{x}+1|+\mathrm{k}$
Question 5.
Match the following.

Answer:
(a) - (iii)
(b) - (iv)
(c) - (i)
(d) - (ii)
Question 6.
If demand function $p=35-2 x-x^2$ and demand $x_0=3$, then market price $p_0$ is
Answer:
$
\begin{aligned}
& \mathrm{p}_0=35-2(3)-3^2 \\
& =35-6-9 \\
& =20
\end{aligned}
$
Question 7.
If supply law is $\mathrm{p}=4-\mathrm{x}+\mathrm{x}^2$ and $\mathrm{p}_0=6$, then demand $\mathrm{x}_0$ is
Answer:
$
\begin{aligned}
& x^2-x+4=6 \\
& \Rightarrow x^2-x-2=0 \\
& \Rightarrow(x-2)(x+1)=0 \\
& \Rightarrow x=2
\end{aligned}
$

Question 8.
Say True or False.
(a) Average cost function $\mathrm{AC}=\frac{\mathrm{MC}}{x}$
(b) Total cost function $\mathrm{C}=\int(\mathrm{MC})^x \mathrm{dx}+\mathrm{k}$
(c) Demand function $=\mathrm{P}=\mathrm{Rx}$
(d) Elasticity of demand $\eta_d=\frac{-p}{x} \frac{d x}{d p}$
(e) Under market equilibrium $x_0=p_0$
Answer:
(a) False
(b) True
(c) False
(d) True
(e) False
II. 2 Mark Questions
Question 1.
Find the area bounded by the curve $y^2=x^3$ and the lines $x=0, y=1$ and $y=2$
Solution:

$
\begin{aligned}
\text { Area } & =\int_1^2 x d y . \text { Here } y^2=x^3 \Rightarrow x=y^{\frac{2}{3}} \\
\text { So Area } & =\int_1^2 \frac{y^{\frac{2}{3}}}{3} d y \\
\text { Area } & =\left(\frac{y^{\frac{5}{3}}}{\frac{5}{3}}\right)_1^2=\frac{3}{5}\left(2^{\frac{5}{3}}-1\right) \text { sq.units }
\end{aligned}
$
Question 2.
Find the area bounded by one arch of the curve $y=\sin$ ax and the $x$-axis.
Solution:

The limits for one arch of the curve $y=\sin a x$ are $x=0$ and $x=\frac{\pi}{a}$
$
\begin{aligned}
\text { So area } & =\int_o^{\frac{\pi}{a}} \sin a x d x=\left[\frac{-\cos a x}{a}\right]_0^{\frac{\pi}{a}}=\frac{-1}{a}(\cos \pi-\cos 0) \\
& =\frac{-1}{a}(-1-1)=\frac{2}{a} \text { sq.units }
\end{aligned}
$

Question 3.
The marginal cost of a commodity is $3 x^2-2 x+8$. If there is no fixed cost find the total cost.
Solution:
$
\begin{aligned}
& C=\int(M C) d x+k \\
& =\int\left(3 x^2-2 x+8\right) d x+k \\
& =x^3-x^2+8 x+k
\end{aligned}
$
No fixed cost $\Rightarrow \mathrm{k}=0$
So total cost, $C=x^3-x^2+8 x$
Question 4.
For the marginal revenue function $M R=3-2 x-x^2$, find the revenue function and demand function.
Solution:
$
\begin{aligned}
\mathrm{R} & =\int\left(3-2 x-x^2\right) d x+k \\
& =3 x-x^2-\frac{x^3}{3}+k
\end{aligned}
$
Since $\mathrm{R}=0$, when $x=0, k=0$
$
\therefore \quad \mathrm{R}=3 x-x^2-\frac{x^3}{3}
$
Demand function $\quad \mathrm{P}=\frac{\mathrm{R}}{x}=3-x-\frac{x^2}{3}$
Question 5.
The marginal cost at a production level of $x$ units is $C^{\prime}(x)=85+\frac{375}{x^2}$. Find the cost of producing 10 extra units, after 15 units have been produced.
Solution:
$
C(x)=\int\left(85+\frac{375}{x^2}\right) d x
$

But 15 units have been produced. We have to find the extra cost for 10 units that is, from 15 to 25 units. Hence
$
\begin{aligned}
C & =\int_{15}^{25}\left(85+\frac{375}{x^2}\right) d x=\left(85 x-\frac{375}{x}\right)_{15}^{25} \\
& =85(25)-\frac{375}{25}-85(15)+\frac{375}{15} \\
& =850-15+25 \\
& =860
\end{aligned}
$
Required cost is ₹ 860
III. 3 and 5 Marks Questions
Question 1.
Find the area under the curve $y=4 x^2-8 x+6$ bounded by the $y$-axis, $x$-axis and the ordinate at $x$ $=2$.
Solution:
The shaded area is given by $\int_0^2 y d x$

$
\begin{aligned}
\text { Area } & =\int_0^2\left(4 x^2-8 x+6\right) d x \\
& =\left(4 \frac{x^3}{3}-\frac{8 x^2}{2}+6 x\right)_0^2=\frac{4}{3}(8)-4(4)-6(2)-0 \\
& =\frac{20}{3} \text { sq.units }
\end{aligned}
$

Question 2.
The marginal revenue for a commodity is $\mathrm{MR}=\frac{e^x}{100}+x+x^2$. Find the revenue function.

Solution:
$
\begin{aligned}
\mathrm{R} & =\int(\mathrm{MR}) d x+k=\int\left(\frac{e^x}{100}+x+x^2\right) d x+k \\
& =\frac{e^x}{100}+\frac{x^2}{2}+\frac{x^3}{3}+k
\end{aligned}
$
When no product is sold, revenue is zero.
$
\Rightarrow x=0, \mathrm{R}=0
$
Thus
$
\begin{aligned}
0 & =\frac{e^0}{100}+0+k \quad \therefore k=-\frac{1}{100} \\
\mathrm{R} & =\frac{e^x}{100}+\frac{x^2}{2}+\frac{x^3}{3}-\frac{1}{100}
\end{aligned}
$
Question 3.
The elasticity of demand with respect to price $\mathrm{p}$ is $\frac{x-5}{x} \mathrm{x}>5$. Find the demand function if the price is 2 when demand is 7 . Also find the revenue function.
Solution:
Elasticity of demand
$
\text { (i.e) } \begin{aligned}
& \frac{-\mathrm{p}}{x} \frac{d x}{d p} & =\frac{x-5}{x} \\
\Rightarrow & \frac{d x}{x-5} & =-\frac{d p}{p}
\end{aligned}
$
Integrating both sides,
$
\int \frac{d x}{x-5}=-\int \frac{d p}{p}+\log k
$

$
\begin{aligned}
& \Rightarrow \log (\mathrm{x}-5)=-\log \mathrm{p}+\log \mathrm{k} \\
& \Rightarrow \log (\mathrm{x}-5)+\log \mathrm{p}=\log \mathrm{k} \\
& \Rightarrow \mathrm{p}(\mathrm{x}-5)=\mathrm{k}
\end{aligned}
$
When $\mathrm{p}=2, \mathrm{x}=1 \Rightarrow \mathrm{k}=4$
Demand function is $\mathrm{P}=\frac{4}{x-5}$
Revenue $\mathrm{R}=\mathrm{px}$ or $\mathrm{R}=\frac{4 x}{x-5}, \mathrm{x}>5$
Question 4.
The marginal cost and marginal revenue of a commodity are given by $C^{\prime}(x)=4+0.08 x$ and $R^{\prime}(x)$ $=12$. Find the total profit, given that the total cost at zero output is zero.
Solution:
Given that, $\mathrm{MC}=4+0.08 \mathrm{x}$
$
\begin{aligned}
& \mathrm{C}(\mathrm{x})=\int(4+0.08 \mathrm{x}) \mathrm{dx}+\mathrm{k}_1 \\
& =4 \mathrm{x}+0.08 \frac{x^2}{2}+\mathrm{k}_1 \\
& =4 \mathrm{x}+0.04 \mathrm{x}^2+\mathrm{k}_1
\end{aligned}
$
But given when $\mathrm{x}=0, \mathrm{C}=0$, So $\mathrm{k}_1=0$
$
C(x)=4 x+0.04 x^2
$
Now given that $\mathrm{MR}=12$
$
\mathrm{R}(\mathrm{x})=\int 12 \mathrm{dx}+\mathrm{k}_2=12 \mathrm{x}+\mathrm{k}_2
$
Revenue $\mathrm{R}=0$ when $\mathrm{x}=0 \Rightarrow \mathrm{k}_2=0$
$
\mathrm{R}(\mathrm{x})=12 \mathrm{x}
$
Total profit function $\mathrm{P}(\mathrm{x})=\mathrm{R}(\mathrm{x})-\mathrm{C}(\mathrm{x})$
$
\begin{aligned}
& =12 \mathrm{x}-4 \mathrm{x}-0.04 \mathrm{x}^2 \\
& =8 \mathrm{x}-0.04 \mathrm{x}^2
\end{aligned}
$
Question 5.
The marginal cost $C^{\prime}(x)$ and marginal revenue $R^{\prime}(x)$ are given by $C^{\prime}(x)=20+\frac{x}{20}$ and $R^{\prime}(x)=30$. The fixed cost is ₹ 200 . Find the maximum profit.

Solution:
Given
$
\begin{aligned}
\mathrm{C}^{\prime}(x) & =20+\frac{x}{20} \\
\therefore \mathrm{C}(x) & =\int\left(20+\frac{x}{20}\right) d x+k_1 \\
& =20 x+\frac{x^2}{40}+k_1
\end{aligned}
$
When quantity produced is zero, the fixed cost is Rs. 200 .
(i.e) $x=0, \mathrm{C}=200 \Rightarrow k_1=200$
cost function $\mathrm{C}(x)=20 x+\frac{x^2}{40}+200$
Given
$
\begin{aligned}
\mathrm{R}^{\prime}(x) & =30 \\
\therefore \mathrm{R}(x) & =\int 30 d x+k_2=30 x+k_2
\end{aligned}
$
when no product is sold, revenue $=0$
(i.e) when $x=0, \mathrm{R}=0$
$\therefore$ Revenue, $\mathrm{R}(x)=30 x$
Profit $\mathrm{P}=\mathrm{R}(x)-\mathrm{C}(x)=30 x-20 x-\frac{x^2}{40}-200$
$
=10 x-\frac{x^2}{40}-200
$
Now $\frac{d p}{d x}=10-\frac{x}{20} ; \frac{d p}{d x}=0 \Rightarrow x=200$
$
\frac{d^2 p}{d x^2}=\frac{-1}{20}<0
$
$\therefore$ Profit is maximum when $x=200$
Maximum profit is $P=2000-\frac{40000}{40}-200$
Profit $=₹ 800$

Question 6.
The marginal cost of producing $x$ units is $C^{\prime}(x)=10.6 x$. The fixed cost is $₹ 50$. The selling price per unit is Rs.5. Find (i) total cost function
(ii) total revenue function and
(iii) profit function.
Solution:
Given $C^{\prime}(x)=10.6 x$
$
\begin{aligned}
& C^{\prime}(\mathrm{x})=\int(10.6 \mathrm{x}) \mathrm{dx}+\mathrm{k} \\
& =10.6 \frac{x^2}{2}+\mathrm{k} \\
& =5.3 \mathrm{x}^2+\mathrm{k}
\end{aligned}
$
The fixed cost is ₹ 50
(i.e.) when $\mathrm{x}=0, \mathrm{c}=50 \Rightarrow \mathrm{k}=50$
Hence cost function $\mathrm{C}=5.3 \mathrm{x}^2+50$
Now, total revenue $=$ number of units sold $\times$ price $/$ unit
Since $x$ be the number of units sold.
The selling price is Rs. 5 per unit.
Revenue $\mathrm{R}(\mathrm{x})=5 \mathrm{x}$
Profit $\mathrm{P}=$ Total revenue $-$ Total cost
$
=5 \mathrm{x}-\left(5.3 \mathrm{x}^2+50\right)
$
$
=5 \mathrm{x}-5.3 \mathrm{x}^2-50
$
Question 7.
The demand and supply functions under pure competition are $p_d=16-x^2$ and $p_s=2 x^2+4$. Find the consumer's surplus and producers' surplus at the market equilibrium price.
Solution:
For market equilibrium, $p_d=p_s$
$
\begin{aligned}
& \Rightarrow \quad 16-x^2=2 x^2+4 \\
& \Rightarrow \quad 3 x^2=12 \Rightarrow x=\pm 2 \quad \text { Since } x \neq-2 \\
& x=2(i . e) \quad x_0=2 \\
& p_0=16-x_0^2=16-4=12 \\
&
\end{aligned}
$

$\begin{aligned}
p_0 x_0 & =24 \\
\text { Consumer's surplus CS } & =\int_0^2\left(16-x^2\right) d x-24 \\
& =\left(16 x-\frac{x^3}{3}\right)_0^2-24=32-\frac{8}{3}-24=\frac{16}{3} \text { units } \\
\text { Producer's surplus PS } & =24-\int_0^2\left(2 x^2+4\right) d x \\
& =24-\left[\frac{2 x^3}{3}+4 x\right]_0^2=24-\left(\frac{16}{3}+8\right)=\frac{32}{3} \text { units }
\end{aligned}$