Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.3

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.3

Question 1. Integrate the following with respect to x: \( \int (e^{x \log a} + e^{a \log a} - e^{a \log x}) dx \)
Answer: We need to integrate the given expression. First, we simplify the terms inside the integral using the logarithm property \( e^{k \log b} = b^k \).
\( \int (e^{x \log a} + e^{a \log a} - e^{a \log x}) dx \)
\( \implies \int (e^{\log a^x} + e^{\log a^a} - e^{\log x^a}) dx \)
\( \implies \int (a^x + a^a - x^a) dx \)
Now, we integrate each term separately. The integral of \( a^x \) is \( \frac{a^x}{\log a} \), the integral of a constant \( a^a \) is \( a^a x \), and the integral of \( x^a \) is \( \frac{x^{a+1}}{a+1} \).
\( \implies \frac{a^x}{\log a} + a^a x - \frac{x^{a+1}}{a+1} + C \)
This gives us the final integrated form of the expression.
In simple words: We first change the terms to make them easier to integrate by using log rules. Then, we find the integral of each part separately and add them up, remembering to include the constant 'C' at the end.

🎯 Exam Tip: Remember the basic integration formulas for exponential functions \( \int a^x dx = \frac{a^x}{\log a} \) and power functions \( \int x^n dx = \frac{x^{n+1}}{n+1} \), as they are key here.

Question 2. Integrate `\( \frac { a^x-e^{x \log b} }{e^{x \log a} b^x} \) ` with respect to x.
Answer: We need to integrate the given fractional expression. We will simplify the expression before integrating.
\( \int \frac { a^x-e^{x \log b} }{e^{x \log a} b^x} dx \)
First, apply the logarithm property \( e^{k \log M} = M^k \) to the terms in the numerator and denominator.
\( \implies \int \frac { a^x-e^{\log b^x} }{e^{\log a^x} b^x} dx \)
\( \implies \int \frac { a^x-b^x }{a^x b^x} dx \)
Now, split the fraction into two simpler terms.
\( \implies \int \left( \frac { a^x }{a^x b^x} - \frac { b^x }{a^x b^x} \right) dx \)
\( \implies \int \left( \frac { 1 }{b^x} - \frac { 1 }{a^x} \right) dx \)
Rewrite these terms using negative exponents to make integration easier.
\( \implies \int (b^{-x} - a^{-x}) dx \)
Now, integrate each term. For \( \int k^{-x} dx \), the integral is \( \frac{k^{-x}}{(-1) \log k} \).
\( \implies \frac{b^{-x}}{(-1) \log b} - \frac{a^{-x}}{(-1) \log a} + C \)
\( \implies - \frac{b^{-x}}{\log b} + \frac{a^{-x}}{\log a} + C \)
This is the final integrated form.
In simple words: First, we change the complicated parts of the fraction using a special log rule. Then, we break the big fraction into two smaller ones and simplify them. After that, we integrate each small part and add 'C' at the end.

🎯 Exam Tip: When dealing with integrals involving exponential and logarithmic terms, always try to simplify the expression using exponent and logarithm rules before integrating.

Question 3. Integrate `\( (e^x + 1)^2 e^x \) ` with respect to x.
Answer: We need to integrate the given expression. First, expand the squared term and then multiply by \( e^x \).
\( \int (e^x + 1)^2 e^x dx \)
Expand the term \( (e^x + 1)^2 \) using the formula \( (A+B)^2 = A^2 + 2AB + B^2 \).
\( \implies \int ((e^x)^2 + 2e^x + 1) e^x dx \)
\( \implies \int (e^{2x} + 2e^x + 1) e^x dx \)
Now, multiply each term inside the parenthesis by \( e^x \).
\( \implies \int (e^{2x} \cdot e^x + 2e^x \cdot e^x + e^x \cdot 1) dx \)
\( \implies \int (e^{3x} + 2e^{2x} + e^x) dx \)
Integrate each term separately. Remember that \( \int e^{kx} dx = \frac{e^{kx}}{k} \).
\( \implies \frac{e^{3x}}{3} + \frac{2e^{2x}}{2} + e^x + C \)
\( \implies \frac{e^{3x}}{3} + e^{2x} + e^x + C \)
This is the final integrated form.
In simple words: First, we open up the bracket that is squared. Then, we multiply everything by \( e^x \). After that, we integrate each part one by one and add the constant 'C'.

🎯 Exam Tip: When integrating expressions with powers of \( e^x \), remember to expand first if there's a binomial square, and use the rule \( \int e^{kx} dx = \frac{e^{kx}}{k} \) for each term.

Question 4. Integrate `\( \frac {e^{3x}-e^{-3x} }{e^{x}} \) ` with respect to x.
Answer: We need to integrate the given fractional expression. We can simplify the fraction by dividing each term in the numerator by the denominator.
\( \int \frac {e^{3x}-e^{-3x} }{e^{x}} dx \)
Split the fraction into two separate terms.
\( \implies \int \left( \frac {e^{3x}}{e^{x}} - \frac {e^{-3x}}{e^{x}} \right) dx \)
Use the exponent rule \( \frac{a^m}{a^n} = a^{m-n} \) to simplify each term.
\( \implies \int (e^{3x-x} - e^{-3x-x}) dx \)
\( \implies \int (e^{2x} - e^{-4x}) dx \)
Now, integrate each term separately. The integral of \( e^{kx} \) is \( \frac{e^{kx}}{k} \).
\( \implies \frac{e^{2x}}{2} - \frac{e^{-4x}}{-4} + C \)
\( \implies \frac{e^{2x}}{2} + \frac{e^{-4x}}{4} + C \)
This is the final integrated form.
In simple words: We first divide each part on top by \( e^x \) to make the expression simpler. Then, we integrate the two new terms separately, remembering the constant 'C' at the end.

🎯 Exam Tip: Always simplify complex fractions before integrating. Dividing terms with exponents often helps convert them into simpler forms for direct integration.

Question 5. Integrate `\( \frac {e^{3x}+e^{5x} }{e^{x}+e^{-x}} \) ` with respect to x.
Answer: We need to integrate the given fractional expression. We will simplify the fraction by factoring terms.
\( \int \frac {e^{3x}+e^{5x} }{e^{x}+e^{-x}} dx \)
Factor out \( e^{3x} \) from the numerator and rewrite \( e^{-x} \) as \( \frac{1}{e^x} \) in the denominator.
\( \implies \int \frac {e^{3x}(1+e^{2x}) }{e^{x}+\frac{1}{e^{x}}} dx \)
Combine the terms in the denominator into a single fraction.
\( \implies \int \frac {e^{3x}(1+e^{2x}) }{\frac{e^{2x}+1}{e^{x}}} dx \)
Now, invert the denominator fraction and multiply.
\( \implies \int e^{3x}(1+e^{2x}) \cdot \frac{e^x}{1+e^{2x}} dx \)
Cancel out the common term \( (1+e^{2x}) \) from the numerator and denominator.
\( \implies \int e^{3x} \cdot e^x dx \)
Use the exponent rule \( a^m \cdot a^n = a^{m+n} \).
\( \implies \int e^{3x+x} dx \)
\( \implies \int e^{4x} dx \)
Integrate the simplified exponential term.
\( \implies \frac{e^{4x}}{4} + C \)
This is the final integrated form. This problem shows how algebraic simplification is vital before integration.
In simple words: First, we take out a common part from the top expression and change the bottom expression to a single fraction. Then, we cancel out matching parts from the top and bottom. This makes the integral much simpler to solve.

🎯 Exam Tip: Look for opportunities to factorize and simplify rational functions (fractions) involving exponents. Often, complex-looking integrals become very simple after algebraic manipulation.

Question 6. Integrate `\( (1 - \frac { 1 }{x^2}) e^{x + \frac { 1 }{x}} \) ` with respect to x.
Answer: We need to integrate this expression using the substitution method.
\( \int \left(1 - \frac { 1 }{x^2}\right) e^{x + \frac { 1 }{x}} dx \)
Let \( t = x + \frac{1}{x} \). This substitution helps simplify the exponent of \( e \).
Next, find the derivative of \( t \) with respect to \( x \).
\( \frac{dt}{dx} = \frac{d}{dx} \left(x + x^{-1}\right) \)
\( \implies \frac{dt}{dx} = 1 - x^{-2} \)
\( \implies \frac{dt}{dx} = 1 - \frac{1}{x^2} \)
From this, we see that \( dt = \left(1 - \frac{1}{x^2}\right) dx \).
Substitute \( t \) and \( dt \) into the original integral.
\( \implies \int e^t dt \)
Integrate \( e^t \) with respect to \( t \).
\( \implies e^t + C \)
Finally, substitute back \( t = x + \frac{1}{x} \) to express the answer in terms of \( x \).
\( \implies e^{x + \frac{1}{x}} + C \)
This is the final integrated form.
In simple words: We use a trick called "substitution" by letting the power of 'e' be a new letter, 't'. When we find the change in 't', it matches the other part of the integral. This makes the integral very easy to solve, and then we put the original expression back in.

🎯 Exam Tip: For integrals of the form \( \int f'(x) e^{f(x)} dx \), always consider substitution with \( t = f(x) \). The derivative of \( f(x) \) often appears as a multiplicative factor.

Question 7. Integrate `\( \frac { 1 }{x(\log x)^2} \) ` with respect to x.
Answer: We need to integrate this expression using the substitution method.
\( \int \frac { 1 }{x(\log x)^2} dx \)
Let \( t = \log x \). This substitution simplifies the denominator.
Next, find the derivative of \( t \) with respect to \( x \).
\( \frac{dt}{dx} = \frac{1}{x} \)
From this, we get \( dt = \frac{1}{x} dx \).
Substitute \( t \) and \( dt \) into the original integral.
\( \implies \int \frac{1}{t^2} dt \)
Rewrite \( \frac{1}{t^2} \) as \( t^{-2} \) to apply the power rule for integration.
\( \implies \int t^{-2} dt \)
Integrate \( t^{-2} \) with respect to \( t \): \( \int t^n dt = \frac{t^{n+1}}{n+1} \).
\( \implies \frac{t^{-2+1}}{-2+1} + C \)
\( \implies \frac{t^{-1}}{-1} + C \)
\( \implies - \frac{1}{t} + C \)
Finally, substitute back \( t = \log x \) to express the answer in terms of \( x \).
\( \implies - \frac{1}{\log x} + C \)
This is the final integrated form. The substitution method is very effective here.
In simple words: We let \( \log x \) be 't'. When we find the change in 't', we see it helps to simplify the integral a lot. After solving the simpler integral, we put \( \log x \) back in place of 't'.

🎯 Exam Tip: When you see \( \log x \) and \( \frac{1}{x} \) together in an integral, it's a strong hint to use substitution with \( t = \log x \).

Question 8. Given `\( f'(x) = e^x \) ` and `\( f(0) = 2 \) `, then find `\( f(x) \) `.
Answer: We are given the derivative of a function \( f'(x) \) and a specific value of the function \( f(x) \) at a certain point. We need to find the original function \( f(x) \).
We have \( f'(x) = e^x \).
To find \( f(x) \), we need to integrate \( f'(x) \) with respect to \( x \).
\( f(x) = \int f'(x) dx \)
\( \implies f(x) = \int e^x dx \)
The integral of \( e^x \) is \( e^x \) itself, plus a constant of integration, \( C \).
\( \implies f(x) = e^x + C \)
Now, we use the given condition, \( f(0) = 2 \), to find the value of \( C \).
Substitute \( x = 0 \) and \( f(x) = 2 \) into the equation for \( f(x) \).
\( 2 = e^0 + C \)
Since \( e^0 = 1 \),
\( 2 = 1 + C \)
Subtract 1 from both sides to find \( C \).
\( \implies C = 2 - 1 \)
\( \implies C = 1 \)
Now, substitute the value of \( C \) back into the function \( f(x) \).
\( \implies f(x) = e^x + 1 \)
This is the specific function that satisfies both the derivative and the initial condition.
In simple words: We start with the given rate of change (derivative) and integrate it to find the general function. Then, we use the given point (where \( x \) is 0 and \( f(x) \) is 2) to find the exact number for the constant 'C', giving us the final specific function.

🎯 Exam Tip: When given a derivative and a point, always integrate first to get the general solution with a constant 'C', then use the given point to solve for 'C'.