Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.10

Evaluate the following:

Question 1.
(i) \( \Gamma (4) \)
Answer: The Gamma function is a special mathematical function. For a positive integer \( n \), the Gamma function \( \Gamma(n) \) is defined as \( (n-1)! \). Here, we need to find \( \Gamma(4) \). Using the definition, we get \( \Gamma(4) = \Gamma(3+1) = 3! \). Calculating the factorial, \( 3! = 3 \times 2 \times 1 = 6 \). So, the value of \( \Gamma(4) \) is 6.
(i) \( \Gamma(4) = \Gamma(3 + 1) = 3! = 6 \)
In simple words: The Gamma function for a whole number is just like a factorial. For \( \Gamma(4) \), it is the same as \( (4-1)! \), which is \( 3! \), and that equals 6.

🎯 Exam Tip: Remember the basic formula \( \Gamma(n) = (n-1)! \) for positive integers \( n \). This helps quickly solve such problems without extensive calculations.

Question 1.
(ii) \( \Gamma \left( \frac{9}{2} \right) \)
Answer: To find the value of \( \Gamma \left( \frac{9}{2} \right) \), we use the recurrence relation for the Gamma function: \( \Gamma(z+1) = z\Gamma(z) \). We can repeatedly apply this formula until we reach \( \Gamma \left( \frac{1}{2} \right) \), which is equal to \( \sqrt{\pi} \). Each step reduces the argument by 1.
(ii) \( \Gamma \left( \frac{9}{2} \right) = \frac{7}{2} \Gamma \left( \frac{7}{2} \right) \)
\( = \frac{7}{2} \times \frac{5}{2} \Gamma \left( \frac{5}{2} \right) \)
\( = \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2} \Gamma \left( \frac{3}{2} \right) \)
\( = \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \Gamma \left( \frac{1}{2} \right) \)
\( = \frac{105}{16} \sqrt{\pi} \)
In simple words: To solve for Gamma of \( \frac{9}{2} \), we use a special rule that helps us reduce the number inside the Gamma function step by step. We keep doing this until we get \( \Gamma \left( \frac{1}{2} \right) \), which is always \( \sqrt{\pi} \). Then, we multiply all the fractions together to get the final answer.

🎯 Exam Tip: For fractional arguments of the Gamma function, repeatedly use the property \( \Gamma(z+1) = z\Gamma(z) \) until the argument becomes \( \frac{1}{2} \), at which point you substitute \( \Gamma \left( \frac{1}{2} \right) = \sqrt{\pi} \).

Question 1.
(iii) \( \int_{0}^{\infty} e^{-mx} x^6 dx \)
Answer: This integral is a type of Gamma integral, which has a standard formula. We know that the definite integral \( \int_{0}^{\infty} x^n e^{-ax} dx \) is equal to \( \frac{n!}{a^{n+1}} \). In our given integral, we have \( n=6 \) and \( a=m \). Substituting these values into the formula gives us the solution.
(iii) We know the formula: \( \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} \)
So, for \( \int_{0}^{\infty} e^{-mx} x^6 dx \), we have \( n=6 \) and \( a=m \).
\( \int_{0}^{\infty} e^{-mx} x^6 dx = \frac{6!}{m^{6+1}} = \frac{6!}{m^7} \)
In simple words: This is a special type of integral that can be solved using a known formula. If you see an integral that looks like \( x \) raised to some power, times \( e \) to the power of negative a number times \( x \), from zero to infinity, the answer is simply the factorial of the power of \( x \), divided by the "a" number raised to one more than that power.

🎯 Exam Tip: Memorize the standard Gamma integral formula \( \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} \) as it frequently appears in calculus and can save significant calculation time.

Question 1.
(iv) \( \int_{0}^{\infty} e^{-4x} x^4 dx \)
Answer: This is another example of a Gamma integral, which can be solved directly using the standard formula. The formula states that \( \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} \). For this specific problem, we can identify \( n=4 \) and \( a=4 \). We substitute these values into the formula and then perform the necessary calculations for the factorial and the power. Finally, simplify the fraction.
(iv) We know that: \( \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} \)
Here, \( n=4 \) and \( a=4 \).
So, \( \int_{0}^{\infty} e^{-4x} x^4 dx = \frac{4!}{4^{4+1}} = \frac{4!}{4^5} \)
\( = \frac{4 \times 3 \times 2 \times 1}{4 \times 4 \times 4 \times 4 \times 4} \)
\( = \frac{24}{1024} \)
\( = \frac{3}{128} \)
In simple words: We use a formula for integrals that go from zero to infinity. Here, \( n \) is 4 (from \( x^4 \)) and \( a \) is 4 (from \( e^{-4x} \)). We put these numbers into the formula \( \frac{n!}{a^{n+1}} \). Then we multiply everything out and simplify the fraction to get the final answer.

🎯 Exam Tip: Always identify the values of 'n' and 'a' correctly from the integral \( \int_{0}^{\infty} x^n e^{-ax} dx \) before applying the formula \( \frac{n!}{a^{n+1}} \). Simplify the resulting fraction carefully.

Question 1.
(v) \( \int_{0}^{\infty} e^{-x/2} x^5 dx \)
Answer: This integral also fits the form of a Gamma integral. We need to evaluate \( \int_{0}^{\infty} e^{-ax} x^n dx \). In this case, comparing it with the given integral, we can see that \( n=5 \) and \( a=\frac{1}{2} \). We then substitute these values into the known formula \( \frac{n!}{a^{n+1}} \) and perform the calculation. Remember that dividing by a fraction is the same as multiplying by its reciprocal, which helps in simplifying the result.
(v) We know that: \( \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} \)
Here, \( n=5 \) and \( a=\frac{1}{2} \).
So, \( \int_{0}^{\infty} e^{-x/2} x^5 dx = \frac{5!}{\left( \frac{1}{2} \right)^{5+1}} \)
\( = \frac{5!}{\left( \frac{1}{2} \right)^6} \)
\( = 5! \times 2^6 \)
\( = 120 \times 64 \)
\( = 7680 \)
In simple words: We use the same formula for this integral. Here, \( n \) is 5 and \( a \) is \( \frac{1}{2} \). We put these into the formula \( \frac{n!}{a^{n+1}} \). Since we are dividing by a fraction, we flip the fraction and multiply instead. Then, we calculate the numbers to get the final answer.

🎯 Exam Tip: Be careful with fractional 'a' values in the denominator. \( \frac{1}{(1/k)^p} \) simplifies to \( k^p \), so ensure you correctly invert and multiply when simplifying the expression.

Question 2. If \( f(x) = \begin{cases} x^{2} e^{-2 x}, & x \geq 0 \\ 0, & \text { otherwise } \end{cases} \), then evaluate \( \int_{0}^{\infty} f(x) dx \)
Answer: We are given a piecewise function \( f(x) \). We need to evaluate its definite integral from \( 0 \) to \( \infty \). Since \( f(x) \) is \( 0 \) for \( x < 0 \), we only need to consider the part of the function where \( x \geq 0 \), which is \( x^2 e^{-2x} \). This simplifies our problem to a standard Gamma integral form. We use the formula \( \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} \), where \( n=2 \) and \( a=2 \). After applying the formula, we perform the arithmetic to get the final result.
Given: \( f(x) = \begin{cases} x^{2} e^{-2 x}, & x \geq 0 \\ 0, & \text { otherwise } \end{cases} \)
To evaluate \( \int_{0}^{\infty} f(x) dx \):
Since \( f(x) = x^2 e^{-2x} \) for \( x \geq 0 \), and \( f(x) = 0 \) for \( x < 0 \), the integral becomes:
\( \int_{0}^{\infty} f(x) dx = \int_{0}^{\infty} x^2 e^{-2x} dx \)
We know the formula: \( \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} \)
Here, \( n=2 \) and \( a=2 \).
So, \( \int_{0}^{\infty} x^2 e^{-2x} dx = \frac{2!}{2^{2+1}} = \frac{2!}{2^3} \)
\( = \frac{2 \times 1}{8} \)
\( = \frac{2}{8} \)
\( = \frac{1}{4} \)
In simple words: The function \( f(x) \) is only useful for values of \( x \) that are zero or positive. So, we only integrate the part \( x^2 e^{-2x} \) from \( 0 \) to infinity. This is a special integral we know how to solve. We use the rule \( \frac{n!}{a^{n+1}} \), where \( n \) is 2 and \( a \) is 2. After calculating, the answer is \( \frac{1}{4} \).

🎯 Exam Tip: When dealing with piecewise functions, always determine the correct limits of integration based on the function's definition. For Gamma integrals, identify 'n' and 'a' precisely before applying the formula.