Exercise 11.4 - Chapter 11 - Integral Calculus - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 11.4
Question 1.
If $f^{\prime}(x)=4 x-5$ and $f(2)=1$; find $f(x)$
Solution:
$
\begin{aligned}
& \int f^{\prime}(x) d x=\int(4 x-5) d x \\
& \mathrm{f}(\mathrm{x})=\frac{4 x^2}{2}-5 \mathrm{x}+\mathrm{c} \\
& \mathrm{f}(\mathrm{x})=2 \mathrm{x}^2-5 \mathrm{x}+\mathrm{c}
\end{aligned}
$
But $\mathrm{f}(2)=1$
$
\begin{aligned}
& 2(2)^2-5(2)+\mathrm{c}=1 \\
& 8-10+\mathrm{c}=1 \\
& \mathrm{c}=3
\end{aligned}
$
Thus, $f(x)=2 x^2-5 x+3$
Question 2 .
If $f^{\prime}(x)=9 x^2-6 x$ and $f(0)=-3$; find $f(x)$
Solution:
$
\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\int\left(9 x^2-6 x\right) d x \\
& \mathrm{f}(\mathrm{x})=\frac{9 x^3}{3}-\frac{6 x^2}{2}+\mathrm{c} \\
& \mathrm{f}(\mathrm{x})=3 \mathrm{x}^3-3 \mathrm{x}^2+\mathrm{c}
\end{aligned}
$
Thus, $f(x)=3 x^3-3 x^2-3$
$
f(x)=3\left(x^3-x^2-1\right)
$
Question 3.
If $f^{\prime}(x)=12 x-6$ and $f(1)=30, f^{\prime}(1)=5$ find $f(x)$
Solution:
$
\begin{aligned}
& f^{\prime}(x)=-6 x+c \\
& f^{\prime}(x)=6 x^2-6 x+c \\
& \text { But } f^{\prime}(1)=5 \\
& 6(1)^2-6(1)+c=5 \\
& c=5 \\
& f^{\prime}(x)=6 x^2-6 x+5 \\
& \int f^{\prime \prime}(x) d x=\int\left(6 x^2-6 x+5\right) d x \\
& f(x)=\frac{6 x^3}{3}-\frac{6 x^2}{2}+5 x+c \\
& f(x)=2 x^3-3 x^2+5 x+c \\
& \text { But } f(1)=30
\end{aligned}
$

$
\begin{aligned}
& 2(1)^3-3(1)^2+5(1)+c=30 \\
& 2-3+5+c=30 \\
& c=26 \\
& f(x)=2 x^3-3 x^2+5 x+26
\end{aligned}
$
Question 4.
A ball is thrown vertically upward from the ground with an initial velocity of $39.2 \mathrm{misec}$. If the onl considered is that attributed to the acceleration due to gravity, find
(i) how long will it take for the ball to strike the ground?
(ii) the speed with which will it strike the ground? and
(iii) how high the ball will rise?
Solution:
We know that
$
\begin{aligned}
\frac{d v}{d t} & =a \\
d v & =a d t \\
\int d v & =a \int d t \\
v & =a t+c
\end{aligned}
$
Given, $t=0 ; v=39.2 \mathrm{~m} / \mathrm{s}$ (initial velocity)
$
c=39.2
$
(i) $\Rightarrow$
$
\begin{aligned}
v & =a t+39.2 \\
\text { Also, } \frac{d x}{d t} & =v \\
d x & =v \mathrm{dt} \\
d x & =(a t+39.2) d t \\
\int d x & =\int(a t+39.2) d t
\end{aligned}
$

$
\begin{aligned}
\Rightarrow \text { from }(3) \Rightarrow \frac{a t^2}{2}+39.2 t & =0 \\
t\left[\frac{a t}{2}+39.2\right] & =0 \\
\frac{a t}{2} & =-39.2 \Rightarrow t=\frac{-39.2 \times 2}{-9.8} \\
t & =8 \mathrm{sec}
\end{aligned}
$
$(t=0$ is not possible)
(but $a=-9.8 \mathrm{~m} / \mathrm{s}$ )
(ii) The speed with which will it strike the ground?
At $\mathrm{t}=8 ;(1) \Rightarrow \mathrm{v}=-9.8(8)+39.2$
$
=-78.4+39.2
$
$
=-39.2
$
$\therefore$ The speed with which the ball will strike the ground is $=39.2 \mathrm{~m} / \mathrm{s}$
(iii) At maximum height $\mathrm{v}=0$
$
\begin{aligned}
& \Rightarrow(2) \Rightarrow-9.8 \mathrm{t}+39.2=0 \\
& \mathrm{t}=4 \mathrm{sec} \\
& \Rightarrow(3) \Rightarrow \mathrm{x}=-\frac{9.8 \times 16}{2}+39.2 \times 4 \\
& =-78.4+156.8=78.4 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
Question 5.
A wound is healing in such a way that $t$ days since Sunday the area of the wound has been decreasing at a rate of $\frac{-6}{(t+2)^2} \mathrm{~cm}^2$ per day where $0<\mathrm{r} \leq 8$. If on Monday the area of the wound was $1.4 \mathrm{~cm}^2$
(i) What was the area of the wound on Sunday?
(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the same rate?
Solution:
Let $A$ be the area of wound at time ' $t$ '.

$
\begin{aligned}
& \text { Given } \frac{d \mathrm{~A}}{d t}=\frac{-6}{(t+2)^2} \\
& \int d \mathrm{~A}=-6 \int \frac{1}{(t+2)^2} d t \\
& \mathrm{~A}=-6\left[\frac{-1}{t+2}\right]+c \\
& \mathrm{~A}=\frac{6}{t+2}+\mathrm{c}
\end{aligned}
$
Hint:
Take $t=0$ on Sunday
$t=1$ on Monday
$t=2$ on Tuesday
and so on ....
By the given, condition area of the wound on Monday is $1.4 \mathrm{~cm}^2$
$
\begin{aligned}
& \Rightarrow \mathrm{A}=1.4 ; \mathrm{t}=1 \\
& \Rightarrow 1.4=\frac{6}{t+2}+\mathrm{c} \\
& 1.4=\frac{6}{3}+\mathrm{c} \\
& \mathrm{c}=-0.6
\end{aligned}
$
$\therefore$ Area of wound at any day.
$
\Rightarrow \mathrm{A}=\frac{6}{t+2}-0.6
$
(i) The area of a wound on Sunday

$
\mathrm{t}=0 \Rightarrow \mathrm{A}=\frac{6}{2}-0.6=3-0.6=2.4 \mathrm{~cm}^2
$
(ii) The area of the wound on Thrusday
$
\mathrm{t}=4 \Rightarrow \mathrm{A}=\frac{6}{6}-0.6=1-0.6=0.4 \mathrm{~cm}^2
$