Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.4

Question 1. If \( f'(x) = 4x - 5 \) and \( f(2) = 1 \), find \( f(x) \)
Answer: To find \( f(x) \) from \( f'(x) \), we need to integrate \( f'(x) \) with respect to \( x \). This process reverses differentiation.
\( \int f'(x) dx = \int (4x - 5) dx \)
\( f(x) = \frac{4x^2}{2} - 5x + c \)
\( f(x) = 2x^2 - 5x + c \)
We are given that \( f(2) = 1 \). We use this to find the constant \( c \).
Substitute \( x = 2 \) into the expression for \( f(x) \):
\( f(2) = 2(2)^2 - 5(2) + c = 1 \)
\( 2(4) - 10 + c = 1 \)
\( 8 - 10 + c = 1 \)
\( -2 + c = 1 \)
\( c = 1 + 2 \)
\( c = 3 \)
Thus, the function \( f(x) \) is \( 2x^2 - 5x + 3 \).
In simple words: First, we integrate the given \( f'(x) \) to get \( f(x) \), which adds a constant \( c \). Then, we use the given value \( f(2) = 1 \) to find out what \( c \) is, and finally write the full \( f(x) \) equation.

🎯 Exam Tip: Remember to always include the constant of integration \( c \) when performing indefinite integration, and use any given boundary conditions (like \( f(2) = 1 \)) to solve for its value. Neglecting \( c \) is a common mistake that loses marks.

Question 2. If \( f'(x) = 9x^2 - 6x \) and \( f(0) = -3 \), find \( f(x) \).
Answer: We need to find \( f(x) \) by integrating \( f'(x) \).
Given \( f'(x) = 9x^2 - 6x \).
So, \( \frac{df(x)}{dx} = 9x^2 - 6x \)
We can write this as \( df(x) = (9x^2 - 6x) dx \)
Integrate both sides:
\( \int df(x) = \int (9x^2 - 6x) dx \)
\( f(x) = \int 9x^2 dx - \int 6x dx \)
\( f(x) = 9 \frac{x^3}{3} - 6 \frac{x^2}{2} + c \)
\( f(x) = 3x^3 - 3x^2 + c \) --- (1)
We are also given that \( f(0) = -3 \). We will substitute \( x = 0 \) into equation (1) to find \( c \).
\( f(0) = 3(0)^3 - 3(0)^2 + c = -3 \)
\( 0 - 0 + c = -3 \)
\( c = -3 \)
Now, substitute the value of \( c \) back into equation (1):
\( f(x) = 3x^3 - 3x^2 - 3 \)
This can also be written by factoring out 3:
\( f(x) = 3(x^3 - x^2 - 1) \). This makes the function easier to analyze.
In simple words: We find \( f(x) \) by integrating \( f'(x) \). This gives us a constant \( c \). We then use the fact that \( f(0) = -3 \) to find the value of \( c \) and complete the equation for \( f(x) \).

🎯 Exam Tip: When integrating polynomials, remember to increase the power of \( x \) by one and divide by the new power. Always simplify the coefficients after integration.

Question 3. If \( f''(x) = 12x - 6 \) and \( f(1) = 30, f'(1) = 5 \), find \( f(x) \)
Answer: To find \( f(x) \) from \( f''(x) \), we need to integrate twice. Each integration will introduce a constant.
Given \( f''(x) = 12x - 6 \).
First, integrate \( f''(x) \) to find \( f'(x) \):
\( f'(x) = \int (12x - 6) dx \)
\( f'(x) = \frac{12x^2}{2} - 6x + c_1 \)
\( f'(x) = 6x^2 - 6x + c_1 \)
We are given that \( f'(1) = 5 \). Use this to find \( c_1 \):
\( f'(1) = 6(1)^2 - 6(1) + c_1 = 5 \)
\( 6 - 6 + c_1 = 5 \)
\( c_1 = 5 \)
So, \( f'(x) = 6x^2 - 6x + 5 \).
Next, integrate \( f'(x) \) to find \( f(x) \):
\( f(x) = \int (6x^2 - 6x + 5) dx \)
\( f(x) = \frac{6x^3}{3} - \frac{6x^2}{2} + 5x + c_2 \)
\( f(x) = 2x^3 - 3x^2 + 5x + c_2 \)
We are given that \( f(1) = 30 \). Use this to find \( c_2 \):
\( f(1) = 2(1)^3 - 3(1)^2 + 5(1) + c_2 = 30 \)
\( 2 - 3 + 5 + c_2 = 30 \)
\( 4 + c_2 = 30 \)
\( c_2 = 30 - 4 \)
\( c_2 = 26 \)
Therefore, the function \( f(x) \) is \( 2x^3 - 3x^2 + 5x + 26 \). This process shows how initial conditions help define the specific function.
In simple words: We integrate twice. First, we integrate \( f''(x) \) to get \( f'(x) \), using \( f'(1)=5 \) to find the first constant. Then, we integrate \( f'(x) \) to get \( f(x) \), using \( f(1)=30 \) to find the second constant.

🎯 Exam Tip: For second-order derivatives, you need to integrate twice and will have two constants of integration (\( c_1 \) and \( c_2 \)). Make sure to use both given initial conditions correctly to solve for each constant.

Question 4. A ball is thrown vertically upward from the ground with an initial velocity of 39.2 m / sec. If the only force 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 it will strike the ground? and
(iii) how high the ball will rise?
Answer: Given initial velocity \( u = 39.2 \text{ m/s} \). Let \( s \) be the distance of the ball from the ground at time \( t \). The acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \). Since the ball is thrown upwards, gravity acts downwards, so we use \( -g \).
The equation for displacement is \( s = ut - \frac{1}{2}gt^2 \).
\( s = 39.2t - \frac{1}{2} \times (9.8)t^2 \)
\( s = 39.2t - 4.9t^2 \)

(i) How long will it take for the ball to strike the ground?
When the ball strikes the ground, its displacement \( s = 0 \).
\( 39.2t - 4.9t^2 = 0 \)
Factor out \( t \):
\( t(39.2 - 4.9t) = 0 \)
This gives two possible solutions: \( t = 0 \) (initial launch) or \( 39.2 - 4.9t = 0 \).
Since \( t \neq 0 \) for striking the ground *after* launch:
\( 4.9t = 39.2 \)
\( t = \frac{39.2}{4.9} \)
\( t = 8 \text{ sec} \). This calculation shows the time it takes for the ball to complete its upward journey and fall back down.

(ii) The speed with which it will strike the ground?
To find the speed, we need the velocity function \( v = u - gt \).
\( v = 39.2 - 9.8t \)
At \( t = 8 \text{ sec} \) (when it strikes the ground):
\( v = 39.2 - 9.8(8) \)
\( v = 39.2 - 78.4 \)
\( v = -39.2 \text{ m/s} \)
The negative sign indicates the direction is downwards. The speed (magnitude of velocity) is \( 39.2 \text{ m/s} \). So, the ball strikes the ground with a speed of \( 39.2 \text{ m/s} \). This demonstrates that the landing speed is equal to the initial launch speed when returning to the same height.

(iii) How high the ball will rise?
At the maximum height, the vertical velocity \( v \) of the ball is \( 0 \).
Using \( v = u - gt \):
\( 0 = 39.2 - 9.8t \)
\( 9.8t = 39.2 \)
\( t = \frac{39.2}{9.8} \)
\( t = 4 \text{ sec} \). This is the time to reach maximum height.
Now, substitute \( t = 4 \) into the displacement equation \( s = 39.2t - 4.9t^2 \):
\( s = 39.2(4) - 4.9(4)^2 \)
\( s = 156.8 - 4.9(16) \)
\( s = 156.8 - 78.4 \)
\( s = 78.4 \text{ m} \). The maximum height reached is \( 78.4 \text{ meters} \).
In simple words: We use physics formulas for motion under gravity. For part (i), we find when the height is zero. For part (ii), we find the speed at that time. For part (iii), we find the maximum height by first finding when the speed becomes zero.

🎯 Exam Tip: Remember to use the correct sign for acceleration due to gravity; it's negative for upward motion and positive for downward motion when defining your coordinate system. Speed is the magnitude of velocity, so always positive.

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 \( -\frac{3}{(t+2)^2} \text{ cm}^2 \) per day. If on Monday the area of the wound was \( 2 \text{ cm}^2 \)
(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?
Answer: Let \( A \) be the area of the wound. Sunday is considered the initial time, so \( t = 0 \) for Sunday. Monday is \( t = 1 \), Tuesday is \( t = 2 \), and so on.
The rate of change of the area is given as \( \frac{dA}{dt} = -\frac{3}{(t+2)^2} \text{ cm}^2/\text{day} \).
To find the area \( A(t) \), we integrate this rate:
\( dA = -\frac{3}{(t+2)^2} dt \)
\( \int dA = \int -3(t+2)^{-2} dt \)
\( A = -3 \frac{(t+2)^{-2+1}}{-2+1} + c \)
\( A = -3 \frac{(t+2)^{-1}}{-1} + c \)
\( A = 3(t+2)^{-1} + c \)
\( A = \frac{3}{t+2} + c \) --- (1)
We know that on Monday (\( t = 1 \)), the area of the wound was \( 2 \text{ cm}^2 \). Use this information to find the constant \( c \).
Substitute \( t = 1 \) and \( A = 2 \) into equation (1):
\( 2 = \frac{3}{1+2} + c \)
\( 2 = \frac{3}{3} + c \)
\( 2 = 1 + c \)
\( c = 2 - 1 \)
\( c = 1 \)
So, the equation for the area of the wound at any time \( t \) is \( A = \frac{3}{t+2} + 1 \) --- (2). This equation precisely models the healing process over time.

(i) What was the area of the wound on Sunday?
For Sunday, \( t = 0 \). Substitute \( t = 0 \) into equation (2):
\( A = \frac{3}{0+2} + 1 \)
\( A = \frac{3}{2} + 1 \)
\( A = 1.5 + 1 \)
\( A = 2.5 \text{ sq.cm} \).
The area of the wound on Sunday was \( 2.5 \text{ cm}^2 \).

(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the same rate?
From Sunday (\( t = 0 \)) to Thursday, there are 4 days. So, for Thursday, \( t = 4 \).
Substitute \( t = 4 \) into equation (2):
\( A = \frac{3}{4+2} + 1 \)
\( A = \frac{3}{6} + 1 \)
\( A = 0.5 + 1 \)
\( A = 1.5 \text{ sq.cm} \).
The anticipated area of the wound on Thursday is \( 1.5 \text{ cm}^2 \).
In simple words: First, we integrate the given rate of healing to find a formula for the wound's area over time, including a constant. We use the area on Monday to find this constant. Then, we use the formula to calculate the area for Sunday (\( t=0 \)) and Thursday (\( t=4 \)).

🎯 Exam Tip: Pay close attention to the definition of \( t=0 \) (e.g., Sunday, initial day). Any other day will be \( t \) days after that initial point. Be careful with negative exponents during integration.