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Choose the correct answer in Exercises 21 and 22.

    Q22.    \int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} equals 

                (A)    \frac{\pi}{6}

                (B)    \frac{\pi}{12}

                (C)    \frac{\pi}{24}

                (D)    \frac{\pi}{4}

Given definite integral \int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}

Consider \int \frac{dx}{4+ 9x^2} = \int \frac{dx}{2^2+(3x)^2}

Now, putting 3x = t

we get, 3dx=dt

Therefore we have, \int \frac{dx}{2^2+(3x)^2} = \frac{1}{3}\int \frac{dt}{2^2+t^2}

= \frac{1}{3}\left ( \frac{1}{2}\tan^{-1}\frac{t}{2} \right ) = \frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )

we have the function of x , as f(x) =\frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )

So, by applying the second fundamental theorem of calculus, we get

\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} = f(\frac{2}{3}) - f(0)

= \frac{1}{6}\tan^{-1}\left ( \frac{3}{2}.\frac{2}{3} \right ) -\frac{1}{6}\tan^{-1}0

= \frac{1}{6}\tan^{-1}1 - 0

= \frac{1}{6}\times \frac{\pi}{4} = \frac{\pi}{24}

Therefore the correct answer is C.

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Choose the correct answer in Exercises 20 and 21.

    Q21.    \int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}

                (A)    \frac{\pi}{3}

                (B)    \frac{2\pi}{3}

                (C)    \frac{\pi}{6}

                (D)    \frac{\pi}{12}

Given definite integral \int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}

Consider \int \frac{dx}{1 +x^2} = \tan^{-1}x

we have then the function of x, as f(x) = \tan^{-1}x

By applying the second fundamental theorem of calculus, we will get

\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2} = f(\sqrt3) - f(1)

= \tan^{-1}\sqrt{3} - \tan^{-1}1

=\frac{\pi}{3} - \frac{\pi}{4}

= \frac{\pi}{12}

Therefore the correct answer is D.

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Divya Prakash Singh

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Evaluate the definite integrals in Exercises 1 to 20.

    Q20.    \int_0^1(xe^x + sin\frac{\pi x}{4})dx

Given integral: \int_0^1(xe^x + sin\frac{\pi x}{4})dx

Consider the integral \int (xe^x + sin\frac{\pi x}{4})dx

can be rewritten as: x\int e^x dx - \int \left \{ \left ( \frac{d}{dx}x \right )\int e^x dx \right \}dx +\left \{ \frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}} \right \}

= xe^x -\int e^x dx -\frac{4\pi}{\pi} \cos \frac{x}{4}

= xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}

 So, we have the function of x, f(x) = xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}

Now, by Second fundamental theorem of calculus, we have

I = f(1) - f(0)

= \left (1.e^t-e^t - \frac{4}{\pi}\cos \frac{\pi}{4} \right ) - \left ( 0.e^0 -e^0 -\frac{4}{\pi}\cos 0 \right )

= e-e -\frac{4}{\pi}\left ( \frac{1}{\sqrt2} \right )+1+\frac{4}{\pi}

= 1+\frac{4}{\pi}- \frac{2\sqrt2}{\pi}

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Divya Prakash Singh

Evaluate the definite integrals in Exercises 1 to 20.

    Q19.    \int_0^2\frac{6x+3}{x^2+ 4}

Given integral: \int_0^2\frac{6x+3}{x^2+ 4}

Consider the integral \int \frac{6x+3}{x^2+ 4}

can be rewritten as: \int \frac{6x+3}{x^2+ 4} = 3\int \frac{2x+1}{x^2+4}dx

= 3\int \frac{2x}{x^2+4}dx +3\int \frac{1}{x^2+4}dx

= 3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}

 So, we have the function of x, f(x) =3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}

Now, by Second fundamental theorem of calculus, we have

I = f(2) - f(0)

= \left \{ 3\log(2^2+4)+\frac{3}{2}\tan^{-1}\left ( \frac{2}{2} \right ) \right \}- \left \{ 3\log(0+4)+\frac{3}{2}\tan^{-1}\left ( \frac{0}{2} \right ) \right \}=3\log 8 +\frac{3}{2}\tan^{-1}1 -3\log 4 -\frac{3}{2}\tan^{-1} 0

=3\log 8 +\frac{3}{2}\times\frac{\pi}{4} -3\log 4 -0

=3\log \frac{8}{4} +\frac{3\pi}{8}

or we have =3\log 2 +\frac{3\pi}{8}

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Evaluate the definite integrals in Exercises 1 to 20.

    Q18.    \int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx

Given integral: \int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx

Consider the integral \int (\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx

can be rewritten as: -\int (\cos^2\frac{x}{2} - \sin^2\frac{x}{2})dx = -\int_0^{\pi} \cos x dx

                                                                                    = \sin x

So, we have the function of x, f(x) =\sin x

Now, by Second fundamental theorem of calculus, we have

I = f(\pi) - f(0)

\Rightarrow \sin \pi - \sin 0 = 0-0 =0

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Evaluate the definite integrals in Exercises 1 to 20.

    Q17.    \int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx

Given integral: \int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx

Consider the integral \int (2\sec^2x + x^3 + 2)dx

\int (2\sec^2x + x^3 + 2)dx = 2\tan x +\frac{x^4}{4}+2x 

So, we have the function of x, f(x) = 2\tan x +\frac{x^4}{4}+2x

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{4}) -f(0)

= \left \{ \left ( 2\tan\frac{\pi}{4}+\frac{1}{4}\left ( \frac{\pi}{4} \right )^4+2\frac{\pi}{4} \right ) - \left ( 2\tan 0 +0 +0 \right ) \right \}

=2\tan\frac{\pi}{4} +\frac{\pi^4}{4^5} +\frac{\pi}{2}

2+\frac{\pi}{2}+\frac{\pi^4}{1024}

 

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Divya Prakash Singh

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Evaluate the definite integrals in Exercises 1 to 20.

    Q16.    \int_1^2\frac{5x^2}{x^2 + 4x +3}

Given integral: I = \int_1^2\frac{5x^2}{x^2 + 4x +3}

So, we can rewrite the integral as;

I = \int_1^2 \frac{5x^2}{x^2 + 4x +3}= \int_1^2 \left ( 5 - \frac{20x+15}{x^2 + 4x +3} \right ) dx

= \int_1^2 5 dx - \int_1^2 \frac{20x+15}{x^2+4x+3}dx

= [5x]_1^2 - \int_1^2 \frac{20x+15}{x^2+4x+3}dx

I = 5-I_1  where I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx     .................(1)

Now, consider I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx

Take numerator 20x+15 = A \frac{d}{dx}\left ( x^2+4x+3 \right )+B

= 2A x+(4A+B)

We now equate the coefficients of x and constant term, we get

A= 10 \and\ B =-25

\Rightarrow I_1 = 10\int_1^2 \frac{2x+4}{x^2+4x+3}dx -25\int_1^2 \frac{dx}{x^2+4x+3}

Now take  denominator x^2+4x+3 = t

Then we have (2x+4)dx =dt

\Rightarrow I_{1} =10\int \frac{dt}{t} -25\int \frac{dx}{(x+2)^2-1^2}

= 10\log t -25\left [ \frac{1}{2}\log\left ( \frac{x+2-1}{x+2+1} \right ) \right ]

=[10\log(x^2+4x+3)]_1^2 -25 \left [ \frac{1}{2}\log\left ( \frac{x+1}{x+3} \right ) \right ]_1^2

= \left [ 10\log15 -10\log 8 \right ] -25 \left [ \frac{1}{2}\log\frac{3}{5} -\frac{1}{2}\log\frac{2}{4} \right ]

= \left [ 10\log5 +10\log3 -10\log4-10\log2 \right ] -\frac{25}{2}\left [ \log3 -\log5-\log2+\log4 \right ]= \left ( 10+\frac{25}{2} \right )\log5 + \left ( -10-\frac{25}{2} \right )\log 4 + \left ( 10-\frac{25}{2} \right )\log 3 + \left ( -10+\frac{25}{2} \right )\log 2= \frac{45}{2}\log5 -\frac{45}{2}\log4 - \frac{5}{2}\log3 +\frac{5}{2}\log2

= \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log \frac{3}{2}

Then substituting the value of I_{1} in equation (1), we get

I= 5 -\left ( \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2} \right )

= 5 -\frac{5}{2}\left ( 9\log\frac{5}{4}-\log\frac{3}{2} \right )

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Evaluate the definite integrals in Exercises 1 to 20.

    Q15.    \int_0^1xe^{x^2}dx

Given integral: \int_0^1xe^{x^2}dx

Consider the integral \int xe^{x^2}dx

Putting x^2 = t which gives, 2xdx =dt

As, x\rightarrow0 ,t \rightarrow0  and  as x\rightarrow1 ,t \rightarrow1.

So, we have now:

\therefore I = \frac{1}{2}\int_0^1 e^t dt 

= \frac{1}{2}\int e^t dt = \frac{1}{2} e^t

So, we have the function of x, f(x) = \frac{1}{2} e^t

Now, by Second fundamental theorem of calculus, we have

I = f(1) -f(0)

= \frac{1}{2}e^1 -\frac{1}{2}e^0 = \frac{1}{2}(e-1)

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Evaluate the definite integrals in Exercises 1 to 20.

    Q14.    \int_0^1\frac{2x+3}{5x^2+1}dx

Given integral: \int_0^1\frac{2x+3}{5x^2+1}dx

Consider the integral \int \frac{2x+3}{5x^2+1}dx

Multiplying by 5 both in numerator and denominator:

\int \frac{2x+3}{5x^2+1}dx = \frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx 

=\frac{1}{5}\int \frac{10x+15}{5x^2+1}dx

= \frac{1}{5} \int \frac{10x}{5x^2+1} dx +3\int \frac{1}{5x^2+1} dx

= \frac{1}{5}\int \frac{10x}{5x^2+1}+3\int \frac{1}{5\left ( x^2+\frac{1}{5} \right )}dx

= \frac{1}{5}\log(5x^2+1) +\frac{3}{5}\times \frac{1}{\frac{1}{\sqrt5}} \tan^{-1}\frac{x}{\frac{1}{\sqrt5}}

= \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )

So, we have the function of x, f(x) = \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )

Now, by Second fundamental theorem of calculus, we have

I = f(1) -f(0)

= \left \{ \frac{1}{5}\log(1+5)+\frac{3}{\sqrt5}\tan^{-1}(\sqrt5) \right \} - \left \{ \frac{1}{5}\log(1)+\frac{3}{\sqrt5}\tan^{-1}(0) \right \}

= \frac{1}{5}\log 6 +\frac{3}{\sqrt 5}\tan^{-1}{\sqrt5}

 

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Evaluate the definite integrals in Exercises 1 to 20.

    Q13.    \int_2^3\frac{xdx}{x^2+1}

Given integral: \int_2^3\frac{xdx}{x^2+1}

Consider the integral \int \frac{xdx}{x^2+1}

\int \frac{xdx}{x^2+1} = \frac{1}{2}\int \frac{2x}{x^2+1}dx =\frac{1}{2}\log(1+x^2)

So, we have the function of x, f(x) =\frac{1}{2}\log(1+x^2)

Now, by Second fundamental theorem of calculus, we have

I = f(3) -f(2)

= \frac{1}{2}\left \{ \log(1+(3)^2)-\log(1+(2)^2) \right \}

= \frac{1}{2}\left \{ \log(10)-\log(5) \right \} = \frac{1}{2}\log\left ( \frac{10}{5} \right ) = \frac{1}{2}\log2

 

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