Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.8

Using Second Fundamental Theorem, Evaluate the Following:

Question 1. \( \int_{0}^{1} e^{2x} dx \)
Answer: To evaluate this definite integral, we first find the antiderivative of \( e^{2x} \). The antiderivative of \( e^{ax} \) is \( \frac{1}{a}e^{ax} \). So, for \( e^{2x} \), the antiderivative is \( \frac{1}{2}e^{2x} \). Now, we apply the limits of integration from 0 to 1.
\( \int_{0}^{1} e^{2x} dx = \left[ \frac{e^{2x}}{2} \right]_{0}^{1} \)
We substitute the upper limit (1) and the lower limit (0) into the antiderivative and subtract the results.
\( = \frac{e^{2(1)}}{2} - \frac{e^{2(0)}}{2} \)
\( = \frac{e^2}{2} - \frac{e^0}{2} \)
Since \( e^0 = 1 \), we have:
\( = \frac{e^2}{2} - \frac{1}{2} \)
\( = \frac{1}{2}(e^2 - 1) \). The second fundamental theorem connects differentiation and integration, allowing us to evaluate definite integrals by finding antiderivatives.
In simple words: First, find the formula that, when you take its derivative, gives you \( e^{2x} \). Then, put the top number (1) into this new formula and subtract what you get when you put the bottom number (0) into it.

🎯 Exam Tip: Remember that \( e^0 \) is always 1, not 0. This is a common mistake in exponential integral problems.

Question 2. \( \int_{0}^{1/4} \sqrt{1 - 4x} dx \)
Answer: We need to evaluate the integral of \( \sqrt{1 - 4x} \). This can be written as \( (1 - 4x)^{1/2} \). We use the substitution method to solve this. Let \( u = 1 - 4x \), then \( du = -4 dx \), which means \( dx = -\frac{1}{4} du \).
We also need to change the limits of integration. When \( x = 0 \), \( u = 1 - 4(0) = 1 \). When \( x = \frac{1}{4} \), \( u = 1 - 4(\frac{1}{4}) = 1 - 1 = 0 \).
So the integral becomes:
\( \int_{1}^{0} u^{1/2} \left( -\frac{1}{4} \right) du \)
\( = -\frac{1}{4} \int_{1}^{0} u^{1/2} du \)
To make the integration limits go from lower to upper, we can switch them and change the sign:
\( = \frac{1}{4} \int_{0}^{1} u^{1/2} du \)
Now, we integrate \( u^{1/2} \) using the power rule, which gives \( \frac{u^{3/2}}{3/2} = \frac{2}{3}u^{3/2} \).
\( = \frac{1}{4} \left[ \frac{2}{3}u^{3/2} \right]_{0}^{1} \)
\( = \frac{1}{4} \times \frac{2}{3} \left[ u^{3/2} \right]_{0}^{1} \)
\( = \frac{1}{6} \left[ (1)^{3/2} - (0)^{3/2} \right] \)
\( = \frac{1}{6} [1 - 0] \)
\( = \frac{1}{6} \). This substitution method helps simplify complex integrals into standard forms.
In simple words: We change the expression under the square root into a simpler variable (like 'u'). Then we integrate this simpler expression. Finally, we put back the original numbers for the limits and calculate the final value.

🎯 Exam Tip: When using substitution in definite integrals, always remember to change the limits of integration to match the new variable (u).

Question 3. \( \int_{0}^{1} \frac{x}{x^2+1} dx \)
Answer: To evaluate this integral, we can use a substitution method. Notice that the derivative of the denominator \( x^2+1 \) is \( 2x \). The numerator has \( x \). So, we can make the numerator look like the derivative of the denominator.
We can rewrite the integral as:
\( = \frac{1}{2} \int_{0}^{1} \frac{2x}{x^2+1} dx \)
Now, let \( u = x^2+1 \). Then \( du = 2x \,dx \).
We also need to change the limits of integration:
When \( x = 0 \), \( u = 0^2+1 = 1 \).
When \( x = 1 \), \( u = 1^2+1 = 2 \).
The integral becomes:
\( = \frac{1}{2} \int_{1}^{2} \frac{1}{u} du \)
The integral of \( \frac{1}{u} \) is \( \log|u| \).
\( = \frac{1}{2} \left[ \log|u| \right]_{1}^{2} \)
Now, we substitute the limits:
\( = \frac{1}{2} (\log|2| - \log|1|) \)
Since \( \log|1| = 0 \):
\( = \frac{1}{2} (\log 2 - 0) \)
\( = \frac{1}{2} \log 2 \). Using logarithms helps solve integrals where the numerator is a multiple of the derivative of the denominator.
In simple words: This integral can be solved by noticing that the top part (x) is almost the derivative of the bottom part (\( x^2+1 \)). We adjust it a bit, then use the natural logarithm to find the answer.

🎯 Exam Tip: Always look for patterns where the numerator is the derivative (or a multiple of the derivative) of the denominator, as these usually integrate to a logarithm.

Question 4. \( \int_{0}^{3} \frac{e^x}{1+e^x} dx \)
Answer: This integral is in a special form where the numerator is the derivative of the denominator. Let \( f(x) = 1+e^x \). Then its derivative \( f'(x) = e^x \).
The integral \( \int \frac{f'(x)}{f(x)} dx \) is equal to \( \log|f(x)| \).
So, we have:
\( \int_{0}^{3} \frac{e^x}{1+e^x} dx = \left[ \log|1+e^x| \right]_{0}^{3} \)
Now, we substitute the upper limit (3) and the lower limit (0):
\( = \log|1+e^3| - \log|1+e^0| \)
Since \( e^0 = 1 \):
\( = \log|1+e^3| - \log|1+1| \)
\( = \log|1+e^3| - \log|2| \)
Using the logarithm property \( \log a - \log b = \log \frac{a}{b} \):
\( = \log \left| \frac{1+e^3}{2} \right| \). This type of integral is very common and relies on recognizing the derivative-denominator relationship.
In simple words: If the top part of the fraction is exactly the derivative of the bottom part, the integral is simply the natural logarithm of the bottom part. Then we plug in the numbers to get the final answer.

🎯 Exam Tip: Master the integral form \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \), as it frequently appears in exams and saves time.

Question 5. \( \int_{0}^{1} xe^{x^2} dx \)
Answer: We can solve this integral using the substitution method. Let \( u = x^2 \). Then, we find the derivative of \( u \) with respect to \( x \): \( du = 2x \,dx \).
From this, we can express \( x \,dx \) as \( \frac{1}{2} du \).
Next, we change the limits of integration:
When \( x = 0 \), \( u = (0)^2 = 0 \).
When \( x = 1 \), \( u = (1)^2 = 1 \).
Now, substitute these into the integral:
\( \int_{0}^{1} e^{x^2} (x \,dx) = \int_{0}^{1} e^u \left( \frac{1}{2} du \right) \)
\( = \frac{1}{2} \int_{0}^{1} e^u du \)
The integral of \( e^u \) is \( e^u \).
\( = \frac{1}{2} \left[ e^u \right]_{0}^{1} \)
Substitute the limits back:
\( = \frac{1}{2} (e^1 - e^0) \)
Since \( e^0 = 1 \):
\( = \frac{1}{2} (e - 1) \). Substitution is a powerful technique for converting integrals into simpler, more recognizable forms.
In simple words: We replace \( x^2 \) with a new variable, 'u'. This makes the integral simpler to solve. After integrating \( e^u \), we put the numbers back to get the final answer.

🎯 Exam Tip: When dealing with \( e^{f(x)} \) terms, always consider substituting \( u = f(x) \) as the first step, as it often leads to a simpler integral.

Question 6. \( \int_{1}^{e} \frac{dx}{x(1+\log x)^3} \)
Answer: We will use substitution for this integral. Let \( f(x) = 1 + \log x \).
Then, the derivative of \( f(x) \) is \( f'(x) = \frac{1}{x} \). So, \( f'(x) dx = \frac{1}{x} dx \).
The integral can be rewritten as \( \int (1+\log x)^{-3} \left( \frac{1}{x} dx \right) \).
This matches the form \( \int [f(x)]^n f'(x) dx \), which integrates to \( \frac{[f(x)]^{n+1}}{n+1} \).
First, change the limits of integration based on \( f(x) = 1 + \log x \):
When \( x = 1 \), \( f(1) = 1 + \log 1 = 1 + 0 = 1 \).
When \( x = e \), \( f(e) = 1 + \log e = 1 + 1 = 2 \) (since \( \log e = 1 \)).
So the integral becomes:
\( \int_{1}^{2} u^{-3} du \)
Now, integrate \( u^{-3} \):
\( = \left[ \frac{u^{-3+1}}{-3+1} \right]_{1}^{2} = \left[ \frac{u^{-2}}{-2} \right]_{1}^{2} \)
\( = -\frac{1}{2} \left[ u^{-2} \right]_{1}^{2} \)
\( = -\frac{1}{2} \left[ \frac{1}{u^2} \right]_{1}^{2} \)
Substitute the new limits:
\( = -\frac{1}{2} \left( \frac{1}{(2)^2} - \frac{1}{(1)^2} \right) \)
\( = -\frac{1}{2} \left( \frac{1}{4} - 1 \right) \)
\( = -\frac{1}{2} \left( \frac{1-4}{4} \right) \)
\( = -\frac{1}{2} \left( -\frac{3}{4} \right) \)
\( = \frac{3}{8} \). Recognizing the derivative of \( \log x \) is key to solving this type of integral.
In simple words: We see that part of the expression is `(1+logx)`, and its derivative is `1/x`. By replacing `(1+logx)` with a new letter and changing the limits, the integral becomes much easier to solve using a basic power rule.

🎯 Exam Tip: When an integral involves \( \log x \), always consider \( u = 1 + \log x \) (or similar) as a substitution, as its derivative \( 1/x \) often appears in the integrand.

Question 7. \( \int_{-1}^{1} \frac{2x+3}{x^2+3x+7} dx \)
Answer: We can solve this integral using the substitution method. Let \( u = x^2+3x+7 \).
Then, find the derivative of \( u \) with respect to \( x \): \( \frac{du}{dx} = 2x+3 \), so \( du = (2x+3) dx \).
Next, we change the limits of integration:
When \( x = -1 \), \( u = (-1)^2+3(-1)+7 = 1 - 3 + 7 = 5 \).
When \( x = 1 \), \( u = (1)^2+3(1)+7 = 1 + 3 + 7 = 11 \).
Substitute these into the integral:
\( \int_{-1}^{1} \frac{2x+3}{x^2+3x+7} dx = \int_{5}^{11} \frac{1}{u} du \)
The integral of \( \frac{1}{u} \) is \( \log|u| \).
\( = \left[ \log|u| \right]_{5}^{11} \)
Substitute the limits:
\( = \log|11| - \log|5| \)
Using the logarithm property \( \log a - \log b = \log \frac{a}{b} \):
\( = \log \left( \frac{11}{5} \right) \). This method is very efficient when the numerator is exactly the derivative of the denominator.
In simple words: We notice that the top part of the fraction, \( 2x+3 \), is the exact derivative of the bottom part, \( x^2+3x+7 \). This means the integral is simply the natural logarithm of the bottom part, evaluated between the given numbers.

🎯 Exam Tip: If the numerator of a rational function is the derivative of its denominator, the integral is always the natural logarithm of the absolute value of the denominator.

Question 8. \( \int_{0}^{\pi/2} \sqrt{1+\cos x} dx \)
Answer: To simplify this integral, we use the trigonometric identity \( 1+\cos x = 2\cos^2\left(\frac{x}{2}\right) \).
Substitute this into the integral:
\( \int_{0}^{\pi/2} \sqrt{2\cos^2\left(\frac{x}{2}\right)} dx \)
\( = \int_{0}^{\pi/2} \sqrt{2} \left| \cos\left(\frac{x}{2}\right) \right| dx \)
For the given limits, \( 0 \leq x \leq \frac{\pi}{2} \), the term \( \frac{x}{2} \) will be in the range \( 0 \leq \frac{x}{2} \leq \frac{\pi}{4} \). In this range, \( \cos\left(\frac{x}{2}\right) \) is positive, so \( \left| \cos\left(\frac{x}{2}\right) \right| = \cos\left(\frac{x}{2}\right) \).
So the integral becomes:
\( = \sqrt{2} \int_{0}^{\pi/2} \cos\left(\frac{x}{2}\right) dx \)
Now, we integrate \( \cos\left(\frac{x}{2}\right) \). The integral of \( \cos(ax) \) is \( \frac{1}{a}\sin(ax) \).
Here \( a = \frac{1}{2} \). So the integral is \( \frac{1}{1/2}\sin\left(\frac{x}{2}\right) = 2\sin\left(\frac{x}{2}\right) \).
\( = \sqrt{2} \left[ 2\sin\left(\frac{x}{2}\right) \right]_{0}^{\pi/2} \)
\( = 2\sqrt{2} \left[ \sin\left(\frac{x}{2}\right) \right]_{0}^{\pi/2} \)
Substitute the limits:
\( = 2\sqrt{2} \left( \sin\left(\frac{\pi/2}{2}\right) - \sin\left(\frac{0}{2}\right) \right) \)
\( = 2\sqrt{2} \left( \sin\left(\frac{\pi}{4}\right) - \sin(0) \right) \)
We know \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \) and \( \sin(0) = 0 \).
\( = 2\sqrt{2} \left( \frac{1}{\sqrt{2}} - 0 \right) \)
\( = 2\sqrt{2} \left( \frac{1}{\sqrt{2}} \right) \)
\( = 2 \). Trigonometric identities are powerful tools for simplifying integrals.
In simple words: We use a special math rule (a trigonometric identity) to change \( 1+\cos x \) into `2cos²(x/2)`. This makes the square root disappear and allows us to easily integrate the cosine function.

🎯 Exam Tip: Memorize key trigonometric identities like \( 1+\cos(2\theta) = 2\cos^2\theta \) (and its variation \( 1+\cos x = 2\cos^2(x/2) \)) as they are often needed to simplify integrals involving square roots of trig functions.

Question 9. \( \int_{1}^{2} \frac{x-1}{x^2} dx \)
Answer: To evaluate this integral, we first simplify the integrand by splitting the fraction.
\( \frac{x-1}{x^2} = \frac{x}{x^2} - \frac{1}{x^2} = \frac{1}{x} - x^{-2} \).
Now, we can integrate each term separately:
\( \int_{1}^{2} \left( \frac{1}{x} - x^{-2} \right) dx \)
The integral of \( \frac{1}{x} \) is \( \log|x| \).
The integral of \( x^{-2} \) is \( \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -\frac{1}{x} \).
So, the antiderivative is \( \log|x| - \left(-\frac{1}{x}\right) = \log|x| + \frac{1}{x} \).
Now, we apply the limits of integration from 1 to 2:
\( = \left[ \log|x| + \frac{1}{x} \right]_{1}^{2} \)
Substitute the upper limit (2) and the lower limit (1):
\( = \left( \log|2| + \frac{1}{2} \right) - \left( \log|1| + \frac{1}{1} \right) \)
Since \( \log|1| = 0 \):
\( = \left( \log 2 + \frac{1}{2} \right) - (0 + 1) \)
\( = \log 2 + \frac{1}{2} - 1 \)
\( = \log 2 - \frac{1}{2} \). Breaking down complex fractions often makes them easier to integrate.
In simple words: First, we break the fraction into two simpler parts: `x/x²` and `1/x²`. Then, we integrate each part separately. Finally, we put in the top and bottom numbers and subtract to find the answer.

🎯 Exam Tip: When the numerator is a polynomial and the denominator is a single term, dividing each term in the numerator by the denominator is often the easiest way to simplify the expression before integrating.

Evaluate the Following

Question 10. \( \int_{1}^{4} f(x) dx \) where \( f(x) = \begin{cases} 4x+3, & 1 \leq x \leq 2 \\ 3x+5, & 2 < x \leq 4 \end{cases} \)
Answer: This is an integral of a piecewise function. We need to split the integral into two parts based on the intervals defined for \( f(x) \).
The integral from 1 to 4 is split at \( x=2 \):
\( \int_{1}^{4} f(x) dx = \int_{1}^{2} (4x+3) dx + \int_{2}^{4} (3x+5) dx \)
First, evaluate the integral of \( (4x+3) \):
\( \int (4x+3) dx = 4\frac{x^2}{2} + 3x = 2x^2 + 3x \).
Now, apply the limits from 1 to 2:
\( \left[ 2x^2 + 3x \right]_{1}^{2} = (2(2)^2 + 3(2)) - (2(1)^2 + 3(1)) \)
\( = (8 + 6) - (2 + 3) = 14 - 5 = 9 \).
Next, evaluate the integral of \( (3x+5) \):
\( \int (3x+5) dx = 3\frac{x^2}{2} + 5x \).
Now, apply the limits from 2 to 4:
\( \left[ \frac{3}{2}x^2 + 5x \right]_{2}^{4} = \left( \frac{3}{2}(4)^2 + 5(4) \right) - \left( \frac{3}{2}(2)^2 + 5(2) \right) \)
\( = \left( \frac{3}{2}(16) + 20 \right) - \left( \frac{3}{2}(4) + 10 \right) \)
\( = (24 + 20) - (6 + 10) = 44 - 16 = 28 \).
Finally, add the results of the two parts:
Total integral \( = 9 + 28 = 37 \). For piecewise functions, it's crucial to correctly identify the integration ranges for each function definition.
In simple words: Because the function `f(x)` changes its formula at \( x=2 \), we break the integral into two separate problems. We solve each part using its correct formula and then add the two results together.

🎯 Exam Tip: Always remember to split the integral at the points where the definition of the piecewise function changes. Carefully apply the correct function for each interval.

Question 11. \( \int_{0}^{2} f(x) dx \) where \( f(x) = \begin{cases} 3-2x-x^2, & x \leq 1 \\ x^2+2x-3, & 1 < x \leq 2 \end{cases} \)
Answer: This is another integral of a piecewise function. We need to split the integral into two parts at \( x=1 \), according to the function's definition.
\( \int_{0}^{2} f(x) dx = \int_{0}^{1} (3-2x-x^2) dx + \int_{1}^{2} (x^2+2x-3) dx \)
First, evaluate the integral of \( (3-2x-x^2) \):
\( \int (3-2x-x^2) dx = 3x - 2\frac{x^2}{2} - \frac{x^3}{3} = 3x - x^2 - \frac{x^3}{3} \).
Now, apply the limits from 0 to 1:
\( \left[ 3x - x^2 - \frac{x^3}{3} \right]_{0}^{1} = \left( 3(1) - (1)^2 - \frac{(1)^3}{3} \right) - \left( 3(0) - (0)^2 - \frac{(0)^3}{3} \right) \)
\( = \left( 3 - 1 - \frac{1}{3} \right) - (0) = 2 - \frac{1}{3} = \frac{6-1}{3} = \frac{5}{3} \).
Next, evaluate the integral of \( (x^2+2x-3) \):
\( \int (x^2+2x-3) dx = \frac{x^3}{3} + 2\frac{x^2}{2} - 3x = \frac{x^3}{3} + x^2 - 3x \).
Now, apply the limits from 1 to 2:
\( \left[ \frac{x^3}{3} + x^2 - 3x \right]_{1}^{2} = \left( \frac{(2)^3}{3} + (2)^2 - 3(2) \right) - \left( \frac{(1)^3}{3} + (1)^2 - 3(1) \right) \)
\( = \left( \frac{8}{3} + 4 - 6 \right) - \left( \frac{1}{3} + 1 - 3 \right) \)
\( = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 2 \right) \)
\( = \left( \frac{8-6}{3} \right) - \left( \frac{1-6}{3} \right) = \frac{2}{3} - \left( -\frac{5}{3} \right) = \frac{2}{3} + \frac{5}{3} = \frac{7}{3} \).
Finally, add the results of the two parts:
Total integral \( = \frac{5}{3} + \frac{7}{3} = \frac{12}{3} = 4 \). Careful calculation of each part and their limits is vital for accuracy.
In simple words: This function also changes its rule at \( x=1 \). So, we do two separate integral problems, one from 0 to 1 using the first rule, and another from 1 to 2 using the second rule. Then we add their answers.

🎯 Exam Tip: Pay close attention to the inequality signs in piecewise function definitions (e.g., \( x \leq 1 \) vs. \( 1 < x \)), as they dictate the exact limits for each part of the integral.

Question 12. \( \int_{-1}^{1} f(x) dx \) where \( f(x) = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases} \)
Answer: This function \( f(x) \) is actually the definition of the absolute value function, \( |x| \). We need to split the integral at \( x=0 \) because the function's definition changes there.
\( \int_{-1}^{1} f(x) dx = \int_{-1}^{0} (-x) dx + \int_{0}^{1} (x) dx \)
First, evaluate the integral of \( (-x) \):
\( \int (-x) dx = -\frac{x^2}{2} \).
Now, apply the limits from -1 to 0:
\( \left[ -\frac{x^2}{2} \right]_{-1}^{0} = \left( -\frac{(0)^2}{2} \right) - \left( -\frac{(-1)^2}{2} \right) \)
\( = (0) - \left( -\frac{1}{2} \right) = \frac{1}{2} \).
Next, evaluate the integral of \( (x) \):
\( \int (x) dx = \frac{x^2}{2} \).
Now, apply the limits from 0 to 1:
\( \left[ \frac{x^2}{2} \right]_{0}^{1} = \left( \frac{(1)^2}{2} \right) - \left( \frac{(0)^2}{2} \right) \)
\( = \frac{1}{2} - 0 = \frac{1}{2} \).
Finally, add the results of the two parts:
Total integral \( = \frac{1}{2} + \frac{1}{2} = 1 \). This integral shows how to handle absolute value functions by splitting the domain.
In simple words: Since the function acts differently for negative and positive numbers, we split the integral at zero. We calculate the integral for the negative part and the positive part separately, then add them together.

🎯 Exam Tip: For integrals involving absolute values or piecewise functions, always split the integral at the points where the function's definition or sign changes.

Question 13. \( f(x) = \begin{cases} cx, & 0
Answer: We are given a piecewise function \( f(x) \) and the value of its definite integral over a certain range. We need to find the constant 'c'.
We know that \( \int_{0}^{1} f(x) dx = 2 \).
Looking at the definition of \( f(x) \), for the interval \( 0 < x < 1 \), \( f(x) = cx \). Outside this interval, \( f(x) = 0 \). So, we only need to integrate \( cx \) from 0 to 1.
\( \int_{0}^{1} cx \,dx = 2 \)
We can pull the constant 'c' out of the integral:
\( c \int_{0}^{1} x \,dx = 2 \)
Now, integrate \( x \): \( \int x \,dx = \frac{x^2}{2} \).
\( c \left[ \frac{x^2}{2} \right]_{0}^{1} = 2 \)
Substitute the limits of integration:
\( c \left( \frac{(1)^2}{2} - \frac{(0)^2}{2} \right) = 2 \)
\( c \left( \frac{1}{2} - 0 \right) = 2 \)
\( c \left( \frac{1}{2} \right) = 2 \)
To solve for \( c \), multiply both sides by 2:
\( c = 4 \). This problem demonstrates how to find an unknown constant in a function when given information about its integral.
In simple words: We are given an integral problem where the answer is known (it's 2). Part of the function has an unknown number 'c'. We solve the integral just like normal, but instead of getting a final number, we get a formula with 'c' in it. Then we set this formula equal to 2 and find 'c'.

🎯 Exam Tip: When finding unknown constants, treat the constant as a multiplier during integration. The integral will yield an expression involving the constant, which you can then solve using the given integral value.