Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.1

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.1

Question 1. Integrate the following with respect to x. \( \sqrt {3x+5 } \)
Answer: We need to integrate the given expression \( \sqrt{3x+5} \) with respect to \( x \). First, we rewrite the square root as a power, so \( \sqrt{3x+5} \) becomes \( (3x+5)^{1/2} \).
\[ \int \sqrt{3x+5} \, dx = \int (3x+5)^{1/2} \, dx \] To integrate this, we use the power rule for integration, \( \int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c \).
Here, \( a = 3 \), \( b = 5 \), and \( n = \frac{1}{2} \).
\[ = \frac{(3x+5)^{1/2+1}}{3(1/2+1)} + c \] \[ = \frac{(3x+5)^{3/2}}{3(3/2)} + c \] \[ = \frac{(3x+5)^{3/2}}{9/2} + c \] \[ = \frac{2}{9}(3x+5)^{3/2} + c \] This is the final integrated form of the expression. This method simplifies the integral of linear functions raised to a power.
In simple words: To find the integral of the square root of \( (3x+5) \), we change the square root to a power of one-half. Then, we use a special integration rule that works for expressions like \( (ax+b)^n \). We add one to the power and divide by the new power multiplied by the number next to \( x \).

🎯 Exam Tip: When integrating functions of the form \( (ax+b)^n \), remember to divide by the derivative of the inner function, \( a \), in addition to dividing by \( (n+1) \).

Question 2. \( (9x^2 – \frac { 4 }{x^2})^2 \)
Answer: We need to integrate the expression \( (9x^2 – \frac { 4 }{x^2})^2 \) with respect to \( x \). First, let's expand the square.
Using the formula \( (A-B)^2 = A^2 - 2AB + B^2 \), where \( A = 9x^2 \) and \( B = \frac{4}{x^2} \).
\[ (9x^2 – \frac { 4 }{x^2})^2 = (9x^2)^2 - 2(9x^2)(\frac{4}{x^2}) + (\frac{4}{x^2})^2 \] \[ = 81x^4 - 2 \cdot 9 \cdot 4 \cdot \frac{x^2}{x^2} + \frac{16}{x^4} \] \[ = 81x^4 - 72 + 16x^{-4} \] Now, we integrate this expanded expression term by term.
\[ \int (81x^4 - 72 + 16x^{-4}) \, dx \] For the first term, \( \int 81x^4 \, dx = 81 \frac{x^{4+1}}{4+1} = 81 \frac{x^5}{5} \).
For the second term, \( \int -72 \, dx = -72x \).
For the third term, \( \int 16x^{-4} \, dx = 16 \frac{x^{-4+1}}{-4+1} = 16 \frac{x^{-3}}{-3} = -\frac{16}{3x^3} \).
Combining these, we get the final integral.
\[ = \frac{81x^5}{5} - 72x - \frac{16}{3x^3} + c \] Expanding the square simplifies the process before integration.
In simple words: First, we open up the bracket by squaring the expression, just like you would square any \( (A-B) \). After expanding, we get three simpler parts. Then, we integrate each part separately by adding 1 to the power of \( x \) and dividing by the new power. Don't forget to add \( +c \) at the end.

🎯 Exam Tip: Always expand algebraic expressions before integrating them, especially binomials raised to a power, to make the integration process simpler using the power rule.

Question 3. \( (3 + x) (2 – 5x) \)
Answer: We need to integrate the expression \( (3 + x) (2 – 5x) \) with respect to \( x \). First, we multiply the two binomials to get a polynomial.
\[ (3 + x) (2 – 5x) = 3(2) + 3(-5x) + x(2) + x(-5x) \] \[ = 6 - 15x + 2x - 5x^2 \] \[ = 6 - 13x - 5x^2 \] Now, we integrate this expanded polynomial term by term.
\[ \int (6 - 13x - 5x^2) \, dx \] For the first term, \( \int 6 \, dx = 6x \).
For the second term, \( \int -13x \, dx = -13 \frac{x^{1+1}}{1+1} = -13 \frac{x^2}{2} \).
For the third term, \( \int -5x^2 \, dx = -5 \frac{x^{2+1}}{2+1} = -5 \frac{x^3}{3} \).
Combining these, we get the final integral.
\[ = 6x - \frac{13x^2}{2} - \frac{5x^3}{3} + c \] Multiplying the terms before integration is a key step here. This ensures all variables are in a power format for easy integration.
In simple words: First, multiply the two parts of the expression together to get a normal polynomial. Then, integrate each part of the polynomial separately. For each \( x \) term, increase its power by one and divide by that new power. Don't forget the constant \( +c \) at the end.

🎯 Exam Tip: Always expand products of polynomials before integrating to convert them into a sum or difference of power functions, which are easier to integrate.

Question 4. \( \sqrt{x} (x^3 – 2x + 3) \)
Answer: We need to integrate the expression \( \sqrt{x} (x^3 – 2x + 3) \) with respect to \( x \). First, we rewrite \( \sqrt{x} \) as \( x^{1/2} \) and then multiply it with each term inside the bracket.
\[ \sqrt{x} (x^3 – 2x + 3) = x^{1/2} (x^3 – 2x + 3) \] \[ = x^{1/2} \cdot x^3 - 2x^{1/2} \cdot x^1 + 3x^{1/2} \] Using the rule \( x^m \cdot x^n = x^{m+n} \):
\[ = x^{1/2+3} - 2x^{1/2+1} + 3x^{1/2} \] \[ = x^{7/2} - 2x^{3/2} + 3x^{1/2} \] Now, we integrate this polynomial term by term.
\[ \int (x^{7/2} - 2x^{3/2} + 3x^{1/2}) \, dx \] For each term, we use the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + c \).
\[ = \frac{x^{7/2+1}}{7/2+1} - 2\frac{x^{3/2+1}}{3/2+1} + 3\frac{x^{1/2+1}}{1/2+1} + c \] \[ = \frac{x^{9/2}}{9/2} - 2\frac{x^{5/2}}{5/2} + 3\frac{x^{3/2}}{3/2} + c \] \[ = \frac{2}{9}x^{9/2} - 2 \cdot \frac{2}{5}x^{5/2} + 3 \cdot \frac{2}{3}x^{3/2} + c \] \[ = \frac{2}{9}x^{9/2} - \frac{4}{5}x^{5/2} + 2x^{3/2} + c \] Converting roots to fractional exponents is crucial for applying the power rule of integration.
In simple words: First, change \( \sqrt{x} \) into \( x \) to the power of one-half. Then, multiply this \( x^{1/2} \) by each part inside the bracket, adding the powers of \( x \). After that, integrate each new part separately by adding 1 to its power and dividing by that new power.

🎯 Exam Tip: Always convert all radical expressions (like square roots) into fractional exponents before integrating, as this allows for direct application of the power rule.

Question 5. \( \frac { 8x+13 }{\sqrt{4x+7}} \)
Answer: We need to integrate the expression \( \frac { 8x+13 }{\sqrt{4x+7}} \) with respect to \( x \). We can rewrite the numerator to involve a multiple of the term inside the square root, \( (4x+7) \).
Let \( 8x+13 = A(4x+7) + B \).
Comparing coefficients of \( x \): \( 8 = 4A \implies A = 2 \).
Comparing constant terms: \( 13 = 7A + B \implies 13 = 7(2) + B \implies 13 = 14 + B \implies B = -1 \).
So, \( 8x+13 = 2(4x+7) - 1 \).
Now substitute this back into the integral:
\[ \int \frac { 2(4x+7) - 1 }{\sqrt{4x+7}} \, dx \] Separate into two fractions:
\[ = \int \left( \frac { 2(4x+7) }{\sqrt{4x+7}} - \frac { 1 }{\sqrt{4x+7}} \right) \, dx \] Rewrite with exponents:
\[ = \int \left( 2(4x+7)^{1/2} - (4x+7)^{-1/2} \right) \, dx \] Now integrate term by term using the rule \( \int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c \).
For the first term, \( 2 \int (4x+7)^{1/2} \, dx = 2 \cdot \frac{(4x+7)^{1/2+1}}{4(1/2+1)} = 2 \cdot \frac{(4x+7)^{3/2}}{4(3/2)} = 2 \cdot \frac{(4x+7)^{3/2}}{6} = \frac{1}{3}(4x+7)^{3/2} \).
For the second term, \( - \int (4x+7)^{-1/2} \, dx = - \frac{(4x+7)^{-1/2+1}}{4(-1/2+1)} = - \frac{(4x+7)^{1/2}}{4(1/2)} = - \frac{(4x+7)^{1/2}}{2} \).
Combine the results:
\[ = \frac{1}{3}(4x+7)^{3/2} - \frac{1}{2}(4x+7)^{1/2} + c \] Manipulating the numerator to match the denominator is a common and effective technique for these types of integrals. This makes it possible to use the power rule for integration.
In simple words: To integrate this fraction, we first change the top part so that it contains a multiple of the bottom part, \( (4x+7) \). This lets us split the fraction into two simpler parts. Then, we change the square roots into powers of one-half. Finally, we integrate each part using the power rule, remembering to divide by the number next to \( x \).

🎯 Exam Tip: For integrals of the form \( \int \frac{ax+b}{\sqrt{cx+d}} \, dx \), try to express the numerator as a linear combination of the denominator or its derivative to simplify the integral into basic power functions.

Question 6. \( \frac { 1 }{\sqrt{x+1}+\sqrt{x-1}} \)
Answer: We need to integrate the expression \( \frac { 1 }{\sqrt{x+1}+\sqrt{x-1}} \) with respect to \( x \). The first step is to rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{x+1} - \sqrt{x-1} \).
\[ \int \frac { 1 }{\sqrt{x+1}+\sqrt{x-1}} \, dx = \int \frac { 1 }{(\sqrt{x+1}+\sqrt{x-1})} \cdot \frac { (\sqrt{x+1}-\sqrt{x-1}) }{(\sqrt{x+1}-\sqrt{x-1})} \, dx \] Using the difference of squares formula \( (A+B)(A-B) = A^2 - B^2 \) in the denominator:
\[ = \int \frac { \sqrt{x+1}-\sqrt{x-1} }{(x+1)-(x-1)} \, dx \] \[ = \int \frac { \sqrt{x+1}-\sqrt{x-1} }{x+1-x+1} \, dx \] \[ = \int \frac { \sqrt{x+1}-\sqrt{x-1} }{2} \, dx \] Now, we can rewrite the square roots as powers and integrate:
\[ = \frac{1}{2} \int \left( (x+1)^{1/2} - (x-1)^{1/2} \right) \, dx \] Integrate term by term using the rule \( \int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c \). Here \( a=1 \) for both terms.
\[ = \frac{1}{2} \left[ \frac{(x+1)^{1/2+1}}{1/2+1} - \frac{(x-1)^{1/2+1}}{1/2+1} \right] + c \] \[ = \frac{1}{2} \left[ \frac{(x+1)^{3/2}}{3/2} - \frac{(x-1)^{3/2}}{3/2} \right] + c \] \[ = \frac{1}{2} \left[ \frac{2}{3}(x+1)^{3/2} - \frac{2}{3}(x-1)^{3/2} \right] + c \] \[ = \frac{1}{3}(x+1)^{3/2} - \frac{1}{3}(x-1)^{3/2} + c \] Rationalizing the denominator transforms a complex fraction into a simpler form that can be integrated using the power rule. This technique is often useful when dealing with sums or differences of square roots in the denominator.
In simple words: First, we remove the square roots from the bottom of the fraction by multiplying the top and bottom by the "opposite" of the bottom part. This process is called rationalizing the denominator. After simplifying, the problem becomes much easier to integrate. We then change the square roots into powers of one-half and integrate each part separately.

🎯 Exam Tip: When faced with an integral involving sums or differences of square roots in the denominator, always try rationalizing the denominator first. This often simplifies the expression significantly for easier integration.

Question 7. Given \( f'(x) = x + b \), \( f(1) = 5 \) and \( f(2) = 13 \), then find \( f(x) \)
Answer: We are given the derivative of a function, \( f'(x) = x + b \), and two conditions: \( f(1) = 5 \) and \( f(2) = 13 \). We need to find the function \( f(x) \).
First, we integrate \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int f'(x) \, dx = \int (x + b) \, dx \] Using the power rule for integration, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \) and \( \int k \, dx = kx \):
\[ f(x) = \frac{x^{1+1}}{1+1} + bx + c \] \[ f(x) = \frac{x^2}{2} + bx + c \] Now, we use the given conditions to find the values of \( b \) and \( c \).
Condition 1: \( f(1) = 5 \)
Substitute \( x = 1 \) into \( f(x) \):
\[ 5 = \frac{(1)^2}{2} + b(1) + c \] \[ 5 = \frac{1}{2} + b + c \] To clear the fraction, multiply by 2:
\[ 10 = 1 + 2b + 2c \] \[ 2b + 2c = 9 \quad \ldots(1) \] Condition 2: \( f(2) = 13 \)
Substitute \( x = 2 \) into \( f(x) \):
\[ 13 = \frac{(2)^2}{2} + b(2) + c \] \[ 13 = \frac{4}{2} + 2b + c \] \[ 13 = 2 + 2b + c \] \[ 2b + c = 11 \quad \ldots(2) \] Now we have a system of two linear equations with two variables:
(1) \( 2b + 2c = 9 \)
(2) \( 2b + c = 11 \)
Subtract equation (2) from equation (1):
\( (2b + 2c) - (2b + c) = 9 - 11 \)
\( c = -2 \)
Substitute \( c = -2 \) into equation (2):
\( 2b + (-2) = 11 \)
\( 2b - 2 = 11 \)
\( 2b = 13 \)
\( b = \frac{13}{2} \)
Finally, substitute the values of \( b \) and \( c \) back into the expression for \( f(x) \):
\[ f(x) = \frac{x^2}{2} + \frac{13}{2}x - 2 \] This process of integrating and then using given points to find constants is fundamental in solving initial value problems. It helps define the unique function from its derivative.
In simple words: First, we integrate the given derivative \( f'(x) \) to find the general function \( f(x) \), which will have a constant \( c \). Then, we use the two given points, \( f(1)=5 \) and \( f(2)=13 \), to create two equations with the unknown values \( b \) and \( c \). We solve these two equations to find \( b \) and \( c \). Finally, we put these values back into \( f(x) \) to get the exact function.

🎯 Exam Tip: Remember that integrating \( f'(x) \) introduces a constant of integration \( c \). You need as many conditions (like \( f(1)=5 \)) as there are unknown constants to solve for them uniquely.

Question 8. Given \( f'(x) = 8x^3 – 2x \) and \( f (2) = 8 \), then find \( f(x) \)
Answer: We are given the derivative of a function, \( f'(x) = 8x^3 – 2x \), and one condition: \( f(2) = 8 \). We need to find the function \( f(x) \).
First, we integrate \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int f'(x) \, dx = \int (8x^3 – 2x) \, dx \] Using the power rule for integration, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \):
\[ f(x) = 8 \frac{x^{3+1}}{3+1} - 2 \frac{x^{1+1}}{1+1} + c \] \[ f(x) = 8 \frac{x^4}{4} - 2 \frac{x^2}{2} + c \] \[ f(x) = 2x^4 - x^2 + c \] Now, we use the given condition \( f(2) = 8 \) to find the value of \( c \).
Substitute \( x = 2 \) into \( f(x) \):
\[ 8 = 2(2)^4 - (2)^2 + c \] \[ 8 = 2(16) - 4 + c \] \[ 8 = 32 - 4 + c \] \[ 8 = 28 + c \] \[ c = 8 - 28 \] \[ c = -20 \] Finally, substitute the value of \( c \) back into the expression for \( f(x) \):
\[ f(x) = 2x^4 - x^2 - 20 \] This method of integrating the derivative and then using a specific point to find the constant is a standard approach to determine the original function. It shows how the initial conditions help pinpoint a unique function from a family of possible functions.
In simple words: First, we integrate the given derivative \( f'(x) \) to find the general function \( f(x) \), which will include a constant \( c \). Then, we use the given point \( f(2)=8 \) by putting \( x=2 \) and \( f(x)=8 \) into our integrated function. This lets us solve for the value of \( c \). Once we have \( c \), we write out the complete function \( f(x) \).

🎯 Exam Tip: Always remember to include the constant of integration \( c \) when performing indefinite integration. Its value is then determined using any given initial conditions for the function.