Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.5

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5

 

Question 1. Integrate the following with respect to x. \( xe^{-x} \)
Answer: To integrate \( xe^{-x} \), we use the integration by parts formula: \( \int udv = uv - \int vdu \). A common way to handle repeated integration by parts is using the tabular method, which involves successive derivatives of 'u' and repeated integrals of 'dv'.
In this problem, we choose \( u = x \) and \( dv = e^{-x} dx \).
First, find the derivative of \( u \) until it becomes zero:
\( u = x \)
\( u' = 1 \)
\( u'' = 0 \)
Next, find the integrals of \( dv \):
\( \int dv = \int e^{-x} dx \implies v = -e^{-x} \)
\( \int v dx = \int (-e^{-x}) dx \implies v_1 = e^{-x} \)
Now, apply the integration by parts formula \( \int udv = uv - u'v_1 + u''v_2 - \dots \):
\( \int xe^{-x} dx = (x)(-e^{-x}) - (1)(e^{-x}) + C \)
\( = -xe^{-x} - e^{-x} + C \)
\( = -e^{-x} (x + 1) + C \)
This method makes integrating products of functions simpler when one function repeatedly differentiates to zero.
In simple words: We used a special math rule called integration by parts. We picked one part to differentiate (x) and another to integrate (\( e^{-x} \)). Then we followed the steps of the rule to get the final answer.

🎯 Exam Tip: When using integration by parts, carefully choose 'u' (the part to differentiate) and 'dv' (the part to integrate). A good general rule is to pick 'u' as the function that becomes simpler when differentiated (like polynomials) and 'dv' as the part that is easily integrable (like \( e^{ax} \) or \( \sin(ax) \)).

 

Question 2. Integrate the following with respect to x. \( x^3e^{3x} \)
Answer: To integrate \( x^3e^{3x} \), we use the integration by parts formula, which for multiple repetitions can be written as \( \int udv = uv - u'v_1 + u''v_2 - u'''v_3 + \dots \).
Here, we choose \( u = x^3 \) and \( dv = e^{3x} dx \).
First, we find the successive derivatives of \( u \):
\( u = x^3 \)
\( u' = 3x^2 \)
\( u'' = 6x \)
\( u''' = 6 \)
\( u'''' = 0 \)
Next, we find the successive integrals of \( dv \):
\( \int dv = \int e^{3x} dx \implies v = \frac{e^{3x}}{3} \)
\( \int v dx = \int \frac{e^{3x}}{3} dx \implies v_1 = \frac{e^{3x}}{9} \)
\( \int v_1 dx = \int \frac{e^{3x}}{9} dx \implies v_2 = \frac{e^{3x}}{27} \)
\( \int v_2 dx = \int \frac{e^{3x}}{27} dx \implies v_3 = \frac{e^{3x}}{81} \)
Now, substitute these into the generalized formula:
\( \int x^3e^{3x} dx = (x^3)\left(\frac{e^{3x}}{3}\right) - (3x^2)\left(\frac{e^{3x}}{9}\right) + (6x)\left(\frac{e^{3x}}{27}\right) - (6)\left(\frac{e^{3x}}{81}\right) + C \)
Factor out \( e^{3x} \):
\( = e^{3x} \left[ \frac{x^3}{3} - \frac{3x^2}{9} + \frac{6x}{27} - \frac{6}{81} \right] + C \)
Simplify the fractions:
\( = e^{3x} \left[ \frac{x^3}{3} - \frac{x^2}{3} + \frac{2x}{9} - \frac{2}{27} \right] + C \)
This systematic method, often called the tabular method, is very efficient for problems requiring repeated integration by parts where one part eventually differentiates to zero.
In simple words: We used a step-by-step method to integrate this problem. We found the derivatives of \( x^3 \) until zero and the integrals of \( e^{3x} \). Then we combined them using a specific pattern to get the final answer.

🎯 Exam Tip: When dealing with \( x^n e^{ax} \) or \( x^n \sin(ax) \), the tabular method for integration by parts is often faster and less error-prone than applying the standard formula repeatedly. Remember to alternate the signs (+, -, +, -) in the formula.

 

Question 3. Integrate the following with respect to x. \( \log x \)
Answer: To integrate \( \log x \), we use integration by parts, treating \( \log x \) as \( \log x \cdot 1 \).
The integration by parts formula is \( \int udv = uv - \int vdu \).
We choose \( u = \log x \) and \( dv = 1 dx \).
First, find the derivative of \( u \):
\( u = \log x \implies du = \frac{1}{x} dx \)
Next, find the integral of \( dv \):
\( dv = 1 dx \implies \int dv = \int 1 dx \implies v = x \)
Now, substitute these into the integration by parts formula:
\( \int \log x dx = (\log x)(x) - \int (x)\left(\frac{1}{x}\right) dx \)
\( = x \log x - \int 1 dx \)
\( = x \log x - x + C \)
We can also factor out \( x \) to get \( = x(\log x - 1) + C \). This integration is a standard result and is often memorized. This method shows how to integrate a single logarithmic function.
In simple words: To integrate just "log x", we imagine it as "log x times 1". We use a rule called integration by parts. We find the derivative of log x and the integral of 1. Then we put these into the rule to get the answer.

🎯 Exam Tip: When integrating functions like \( \log x \) or \( \arctan x \) which don't seem to have a 'dv' part, always consider '1' as 'dv'. Remember the 'ILATE' rule for choosing 'u': Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. Logarithmic functions (like \( \log x \)) are usually chosen as 'u' because they simplify upon differentiation.

 

Question 4. Integrate the following with respect to x. \( x \log x \)
Answer: To integrate \( x \log x \), we use integration by parts, \( \int udv = uv - \int vdu \).
Following the 'ILATE' rule, we choose \( u = \log x \) (logarithmic comes before algebraic) and \( dv = x dx \).
First, find the derivative of \( u \):
\( u = \log x \implies du = \frac{1}{x} dx \)
Next, find the integral of \( dv \):
\( dv = x dx \implies \int dv = \int x dx \implies v = \frac{x^2}{2} \)
Now, substitute these into the integration by parts formula:
\( \int x \log x dx = (\log x)\left(\frac{x^2}{2}\right) - \int \left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) dx \)
\( = \frac{x^2}{2} \log x - \int \frac{x}{2} dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2} \int x dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2}\left(\frac{x^2}{2}\right) + C \)
\( = \frac{x^2}{2} \log x - \frac{x^2}{4} + C \)
We can also factor out \( \frac{x^2}{2} \) to get \( = \frac{x^2}{2} \left[ \log x - \frac{1}{2} \right] + C \). This method shows a common application of integration by parts for products of different types of functions.
In simple words: We used the integration by parts rule again. We picked log x as 'u' and x as 'dv' because log functions are differentiated first. Then we did the steps of differentiation and integration and put them into the formula to get the answer.

🎯 Exam Tip: For products of algebraic and logarithmic functions, always choose the logarithmic function as 'u' because its derivative simplifies, making the subsequent integral easier to solve. Carefully apply the power rule for integration (\( \int x^n dx = \frac{x^{n+1}}{n+1} \)).

 

Question 5. Integrate the following with respect to x. \( x^n \log x \)
Answer: To integrate \( x^n \log x \), we use integration by parts, \( \int udv = uv - \int vdu \).
Following the 'ILATE' rule, we choose \( u = \log x \) and \( dv = x^n dx \).
First, find the derivative of \( u \):
\( u = \log x \implies du = \frac{1}{x} dx \)
Next, find the integral of \( dv \):
\( dv = x^n dx \implies \int dv = \int x^n dx \implies v = \frac{x^{n+1}}{n+1} \)
Now, substitute these into the integration by parts formula:
\( \int x^n \log x dx = (\log x)\left(\frac{x^{n+1}}{n+1}\right) - \int \left(\frac{x^{n+1}}{n+1}\right)\left(\frac{1}{x}\right) dx \)
\( = \frac{x^{n+1}}{n+1} \log x - \int \frac{x^{n+1-1}}{n+1} dx \)
\( = \frac{x^{n+1}}{n+1} \log x - \frac{1}{n+1} \int x^n dx \)
\( = \frac{x^{n+1}}{n+1} \log x - \frac{1}{n+1}\left(\frac{x^{n+1}}{n+1}\right) + C \)
\( = \frac{x^{n+1}}{n+1} \log x - \frac{x^{n+1}}{(n+1)^2} + C \)
We can also factor out \( \frac{x^{n+1}}{n+1} \) to get \( = \frac{x^{n+1}}{n+1} \left[ \log x - \frac{1}{n+1} \right] + C \). This result is a generalized form for integrating products of a power function and a logarithm.
In simple words: We used the integration by parts rule. We chose log x as 'u' and \( x^n \) as 'dv'. We found their derivatives and integrals and then put them into the formula to solve. The \( n \) here is just a general number.

🎯 Exam Tip: Always remember that \( \int x^n dx = \frac{x^{n+1}}{n+1} \) is valid for \( n \neq -1 \). When \( n = -1 \), the integral is \( \int \frac{1}{x} dx = \log |x| \). Be careful with the algebraic simplification of exponents (e.g., \( x^{n+1} \cdot x^{-1} = x^n \)).

 

Question 6. Integrate the following with respect to x. \( x^5 e^{x^2} \)
Answer: To integrate \( x^5 e^{x^2} \), we first use a substitution to simplify the exponent of \( e \).
Let \( t = x^2 \).
Then, \( dt = 2x dx \implies x dx = \frac{dt}{2} \).
Now, rewrite the integral in terms of \( t \). We need \( x^5 \). Since \( t = x^2 \), then \( x^4 = t^2 \).
So, \( \int x^5 e^{x^2} dx = \int x^4 \cdot x \cdot e^{x^2} dx \)
\( = \int t^2 \cdot e^t \cdot \frac{dt}{2} \)
\( = \frac{1}{2} \int t^2 e^t dt \)
Now, we integrate \( \int t^2 e^t dt \) using the tabular method for integration by parts.

Applying the formula \( \int udv = uv - u'v_1 + u''v_2 - u'''v_3 + \dots \):
\( \int t^2 e^t dt = (t^2)(e^t) - (2t)(e^t) + (2)(e^t) + C_1 \)
\( = e^t (t^2 - 2t + 2) + C_1 \)
Now, substitute this back into the original expression with the \( \frac{1}{2} \) factor:
\( \int x^5 e^{x^2} dx = \frac{1}{2} [e^t (t^2 - 2t + 2)] + C \)
Finally, replace \( t \) with \( x^2 \):
\( = \frac{1}{2} e^{x^2} ((x^2)^2 - 2(x^2) + 2) + C \)
\( = \frac{1}{2} e^{x^2} (x^4 - 2x^2 + 2) + C \)
This problem demonstrates combining substitution with integration by parts, which is a powerful technique for complex integrals. The initial substitution transforms the integral into a more manageable form.
In simple words: First, we changed \( x^2 \) to a new letter, 't', to make the problem easier. This also changed \( dx \) to \( dt \). After the change, we used the integration by parts rule, similar to before, to solve it. In the end, we put \( x^2 \) back where 't' was to get the final answer in terms of x.

🎯 Exam Tip: When an integral involves \( e^{f(x)} \) or \( \sin(f(x)) \), consider substituting \( u = f(x) \) first. If the derivative of \( f(x) \) (or a part of it) is also present in the integral, the substitution simplifies the expression significantly before applying other integration techniques like integration by parts.