12th NCERT Integral Exercise 7.1 Questions 22
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Find antiderivative of the following functions by method of inspection

Antiderivative of function is a function whose derivative will be given function

Question (1)

sin 2x
Solution:
Antiderivative of sin2x is the function whose derivative will be sin2x
$\begin{array}{l}\;\frac{d}{{dx}}\left( {\cos 2x} \right) = - \sin 2x \cdot 2\\ \Rightarrow \sin 2x = - \frac{1}{2}\frac{d}{{dx}}\left( {\cos 2x} \right)\\\sin 2x = \frac{d}{{dx}}\left( { - \frac{1}{2}\cos 2x} \right)\end{array}$ ∴ Antiderivative of sin2x is ${ - \frac{1}{2}\cos 2x}$

Question (2)

cos3x
Solution:
$\begin{array}{l}\;\frac{d}{{dx}}\left( {\sin 3x} \right) = \cos 3x \cdot 3\\ \Rightarrow \cos 3x = \frac{1}{3}\frac{d}{{dx}}\left( {\sin 3x} \right)\\\cos 3x = \frac{d}{{dx}}\left( {\frac{1}{3}\sin 3x} \right);Antiderivative\;\cos 3x\;is\;\frac{1}{3}\sin 3x\end{array}$

Question (3)

e2x
Solution:
$\begin{array}{l}\frac{d}{{dx}}\left( {{e^{2x}}} \right) = 2 \cdot {e^{2x}}\\ \Rightarrow {e^{2x}} = \frac{1}{2}\frac{d}{{dx}}{e^{2x}}\\{e^{2x}} = \frac{d}{{dx}}\left( {\frac{1}{2}{e^{2x}}} \right)\\Antiderivative\;of\;{e^{2x}}\;is\;\frac{1}{2}{e^{2x}}\end{array}$

Question (4)

(ax+b)2
Solution:
Antiderivative of (ax+b)2 is a function whose derivative is (ax+b)2
$\begin{array}{l}\frac{d}{{dx}}{\left( {ax + b} \right)^3} = 3 \cdot {\left( {ax + b} \right)^2} \times a\\ \Rightarrow {\left( {ax + b} \right)^2} = \frac{1}{{3a}}\frac{d}{{dx}}{\left( {ax + b} \right)^3}\\{\left( {ax + b} \right)^2} = \frac{d}{{dx}}\left[ {\frac{{{{\left( {ax + b} \right)}^3}}}{{3a}}} \right]\end{array}$
Antiderivative is ${\frac{{{{\left( {ax + b} \right)}^3}}}{{3a}}}$

Question (5)

sin2x-4e2x
Solution:
Antiderivative of sin2x-4e2x is a function whose derivative is sin2x-4e2x
$\begin{array}{l}\frac{d}{{dx}}\left( { - \frac{1}{2}\cos 2x - \frac{4}{3}{e^{3x}}} \right) = \frac{1}{2}\frac{d}{{dx}}\cos 2x - \frac{4}{3}\frac{d}{{dx}}{e^{3x}}\\\frac{d}{{dx}}\left( { - \frac{1}{2}\cos 2x - \frac{4}{3}{e^{3x}}} \right) = \frac{1}{2}\sin 2x \cdot 2 - \frac{4}{3}{e^{3x}} \cdot 3\\\frac{d}{{dx}}\left( { - \frac{1}{2}\cos 2x - \frac{4}{3}{e^{3x}}} \right) = \left( {\sin 2x - 4{e^{3x}}} \right)\\\sin 2x - 4{e^{3x}} = \frac{d}{{dx}}\left( {\frac{1}{2}\cos 2x - \frac{4}{3}{e^{3x}}} \right)\\\end{array}$ Antiderivative ${\frac{1}{2}\cos 2x - \frac{4}{3}{e^{3x}}}$

Find the following integrals in Exercise 6 to 20

Question (6)

$\int {\left( {4{e^{3x}} + 1} \right)dx}$ Solution:
$\begin{array}{l}I = \;\int {\left( {4{e^{3x}} + 1} \right)dx} \\I = \int {4{e^{3x}}dx + \int {1dx} } \\I = 4\int {{e^{3x}}dx + x + c\;} \left( {\int {dx = x} } \right)\end{array}$ Let 3x=t
differentiate w.r.t x
$\begin{array}{l}3 = \frac{{dt}}{{dx}}\\3dx = dt\\dx = \frac{{dt}}{3}\end{array}$ Replacing value of t and dx $\begin{array}{l}I = 4\int {{e^t}\frac{{dt}}{3}} + x + c\\I = \frac{4}{3}\int {{e^t}dt} + x + c\\I = \frac{4}{3}{e^t} + x + c\end{array}$ Replace value of t
$I = \frac{4}{3}{e^{3x}} + x + c$

Question (7)

$\int {{x^2}\left( {1 - \frac{1}{{{x^2}}}} \right)dx}$
Solution:
In Integration there is no formula for multiplication. Convert multiplication in to addition or subtraction
$\begin{array}{l}I = \int {{x^2}\left( {1 - \frac{1}{{{x^2}}}} \right)dx} \\I = \int {\left( {{x^2} - \frac{{{x^2}}}{{{x^2}}}} \right)dx} \\I = \int {{x^2}dx - \int {1dx} } \\I = \frac{{{x^3}}}{3} - x + c\;\left( {\int {{x^n} = \frac{{{x^{n + 1}}}}{{n + 1}}} } \right)\end{array}$

Question (8)

$\int {\left( {a{x^2} + bx + c} \right)dx}$ Solution:
$\begin{array}{l}I = \int {\left( {a{x^2} + bx + c} \right)dx} \\I = \int {a{x^2} + \int {bxdx + } } \int {cdx} \\I = a\int {{x^2} + b\int {xdx + } } c\int {dx} \\I = a\frac{{{x^3}}}{3} + b\frac{{{x^2}}}{2} + cx + k\end{array}$ Since coefficient of terms can be taken out of integration. Already "c" is there so constant k is used

Question (9)

$\int {\left( {2{x^2} + {e^x}} \right)} dx$ Solution:
$\begin{array}{l}I = \int {\left( {2{x^2} + {e^x}} \right)} dx\\I = \int {2{x^2}} dx + \int {{e^x}dx} \\I = 2\int {{x^2}} dx + \int {{e^x}dx} \end{array}$
Formula used:
$\begin{array}{l} \circ \int {{x^n}dx} = \frac{{{x^n}}}{{n + 1}} + c\\ \circ \int {{e^x}} dx = {e^x} + c\end{array}$
$I = \frac{2}{3}{x^3} + {e^x} + c$

Question (10):

${\int {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)} ^2}dx$ Solution:
First expand by formula
(a-b)2 = a2-2ab+b2
$\begin{array}{l}I = {\int {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)} ^2}dx\\I = \int {\left[ {{{\left( {\sqrt x } \right)}^2} - 2\sqrt x \frac{1}{{\sqrt x }} + {{\left( {\frac{1}{{\sqrt x }}} \right)}^2}} \right]dx} \\I = \int {xdx} - \int {2dx + \int {\frac{1}{x}} } dx\end{array}$
Use following formula:
$\begin{array}{l} \circ \int {{x^n} = \frac{{{x^{n + 1}}}}{{n + 1}}} + c\\ \circ \int {\frac{1}{x}dx = \log \left| x \right|} + c\end{array}$
$I = \frac{{{x^2}}}{2} - 2x + \log \left| x \right| + c$

Question (11):

$\int {\frac{{{x^3} + 5{x^2} - 4}}{{{x^2}}}dx}$
Solution:
There is no rule for division of integration.
Denomeneter is monomial so we can separate terms by dividing monomial to each term.
$\begin{array}{l}I = \int {\frac{{{x^3} + 5{x^2} - 4}}{{{x^2}}}dx} \\I = \left( {\int {\frac{{{x^3}}}{{{x^2}}} + \frac{{5{x^2}}}{{{x^2}}} - \frac{4}{{{x^2}}}} } \right)dx\\I = \left( {\int {x + 5 - \frac{4}{{{x^2}}}} } \right)dx\end{array}$
Use formula: $\frac{1}{{{x^2}}} = {x^{ - 2}}$ and $\int {{x^n} = \frac{{{x^{n + 1}}}}{{n + 1}}} + c$
$\begin{array}{l}I = \int {xdx + \int {5dx - \int {4{x^{ - 2}}} } } dx\\I = \frac{{{x^2}}}{2} + 5x - 4\frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}\\I = \frac{{{x^2}}}{2} + 5x + 4{x^{ - 1}}\\I = \frac{{{x^2}}}{2} + 5x + \frac{4}{x}+c\end{array}$

Question (12):

$\int {\frac{{{x^3} + 3x + 4}}{{\sqrt x }}} dx$ Solution:
As denominator is monomial , divide each term in numerator for separation.
Then use rules of exponents. $\begin{array}{l} \circ \frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\\ \circ \frac{1}{{{a^m}}} = {a^{ - m}}\end{array}$
$\begin{array}{l}I = \int {\frac{{{x^3} + 3x + 4}}{{\sqrt x }}} dx\\I = \int {\left( {\frac{{{x^2}}}{{\sqrt x }} + \frac{{3x}}{{\sqrt x }} + \frac{4}{{\sqrt x }}} \right)} dx\\I = \int {\left( {{x^{3 - \frac{1}{2}}} + 3{x^{1 - \frac{1}{2}}} + 4{x^{\frac{{ - 1}}{2}}}} \right)dx} \\I = \int {{x^{\frac{5}{2}}}dx + 3\int {{x^{\frac{1}{2}}}} dx + } 4\int {{x^{\frac{{ - 1}}{2}}}dx} \\I = \frac{{{x^{\frac{5}{2} + 1}}}}{{\frac{5}{2} + 1}} + 3\frac{{{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} + 4\frac{{{x^{\frac{{ - 1}}{2} + 1}}}}{{\frac{{ - 1}}{2} + 1}} + c\\I = \frac{{{x^{\frac{7}{2}}}}}{{\frac{7}{2}}} + 3\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} + 4\frac{{{x^{\frac{1}{2}}}}}{{\frac{1}{2}}} + c\\I = \frac{2}{7}{x^{\frac{7}{2}}} + 2{x^{\frac{3}{2}}} + 8{x^{\frac{1}{2}}} + c\end{array}$

Question (13):

$\int {\frac{{{x^3} - {x^2} + x - 1}}{{x - 1}}}$
The denominator is binomial. So we can not divide by it to every term, as it will not simplify it.
Factors of numerator
x3-x2+x+1 = x2(x-1) + 1(x+1)
=(x-1)(x2+1)
Now denominator (x-1) will cancel numerator (x-1) factor. It will reduce equation to simple form
$\begin{array}{l}I = \int {\frac{{{x^3} - {x^2} + x - 1}}{{x - 1}}} dx\\I = \int {\frac{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}{{x - 1}}dx} \\I = \int {\left( {{x^2} + 1} \right)dx} \\I = \int {{x^2}dx} + \int {dx} \\I = \frac{{{x^3}}}{3} + x + c\end{array}$

Question (14):

$\int {\left( {1 - x} \right)\sqrt x dx}$ Solution:
First multiply use am×an= am+n
Then formula$\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}}}$
$\begin{array}{l}I = \int {\left( {1 - x} \right)\sqrt x dx} \\I = \int {\sqrt x dx - \int {{x^{\frac{3}{2}}}} dx} \\I = \frac{{{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} - \frac{{{x^{\frac{3}{2} + 1}}}}{{\frac{3}{2} + 1}} + c\\I = \frac{2}{3}{x^{\frac{3}{2}}} - \frac{2}{5}{x^{\frac{5}{2}}} + c\end{array}$

Question (15):

$\int {\sqrt x \left( {3{x^2} + 2x + 3} \right)} dx$ Solution:
$\begin{array}{l}I = \int {\sqrt x \left( {3{x^2} + 2x + 3} \right)} dx\\I = \left( {\int {3{x^2}} \sqrt x + 1x\sqrt x + 3\sqrt x } \right)\\I = \int {\left( {3{x^{\frac{5}{2}}} + 2{x^{\frac{3}{2}}} + 3{x^{\frac{1}{2}}}} \right)} dx\\I = 3\int {{x^{\frac{5}{2}}}dx + 2\int {{x^{\frac{3}{2}}}dx + 3\int {{x^{\frac{1}{2}}}} } } dx\\I = 3 \cdot \frac{{{x^{\frac{7}{2}}}}}{{\frac{7}{2}}} + 2 \cdot \frac{{{x^{\frac{5}{2}}}}}{{\frac{5}{2}}} + 3 \cdot \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} + c\\I = \frac{6}{7}{x^{\frac{7}{2}}} + \frac{4}{5}{x^{\frac{5}{2}}} + 2{x^{\frac{3}{2}}} + c\end{array}$

Question (16):

$\int {\left( {2x - 3\cos x + {e^x}} \right)} dx$ Solution : $\begin{array}{l}I = \int {\left( {2x - 3\cos x + {e^x}} \right)} dx\\I = 2\int {xdx} - 3\int {\cos xdx + \int {{e^x}} } dx\end{array}$
Formula $\begin{array}{l} \circ \int {{e^x}} dx = {e^x} + c\\ \circ \int {\cos x = \sin x + c} \end{array}$
$\begin{array}{l}I = 2\frac{{{x^2}}}{2} - 3\sin x + {e^x} + c\\I = {x^2} - 3\sin x + {e^x} + c\end{array}$

Question (17):

$\int {\left( {2{x^2} - 3\sin x + 5\sqrt x } \right)} dx$ Solution: $\begin{array}{l}I = \int {\left( {2{x^2} - 3\sin x + 5\sqrt x } \right)} dx\\I = 2\int {{x^2}dx - 3\int {\sin xdx + 5\int {{x^{\frac{1}{2}}}dx} } } \\I = 2\frac{{{x^3}}}{3} - 3\left( { - \cos x} \right) + 5\frac{{{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} + c\\I = \frac{2}{3}{x^3} + 3\cos x + \frac{{10}}{3}{x^{\frac{3}{2}}} + c\end{array}$

Question (18):

$\int {\sec x\left( {\sec x + \tan x} \right)dx}$ Solution: $\begin{array}{l}I = \int {\sec x\left( {\sec x + \tan x} \right)dx} \\I = \int {\left( {{{\sec }^2}x + \sec x\tan x} \right)} dx\\I = \int {{{\sec }^2}xdx + \int {\sec x\tan xdx} } \end{array}$
Formula $\begin{array}{l} \circ \int {{{\sec }^2}} xdx = \tan x + c\\ \circ \int {\sec x\tan xdx} = \sec x + c\end{array}$
I = tanx + secx + c

Question (19):

$\int {\frac{{{{\sec }^2}x}}{{\cos e{c^2}x}}} dx$ Solution:
There is no formula for division, we will try to simplify on the basis of trigo. formulas $\begin{array}{l}\sec x = \frac{1}{{\cos x}}\\\frac{1}{{\cos ecx}} = \sin x\end{array}$
$\begin{array}{l}I = \int {\frac{{{{\sec }^2}x}}{{\cos e{c^2}x}}} dx\\I = \int {\frac{1}{{{{\cos }^2}x}}} \times {\sin ^2}xdx\\I = \int {{{\tan }^2}} xdx\end{array}$
There is no formula for ∫tan2x
We can use trigonometric identity tan2x = sec2x - 1
∫sec2xdx= tanx
$\begin{array}{l}I = \int {\left( {{{\sec }^2}x - 1} \right)dx} \\I = \int {{{\sec }^2}xdx - \int {dx} } \\I = \tan x - x + c\end{array}$

Question (20):

$\int {\frac{{2 - 3\sin x}}{{{{\cos }^2}x}}dx}$ Solution:
Denominator is monomial by separating and dividing each term of numerator by denominator we get
$\begin{array}{l}I = \int {\frac{{2 - 3\sin x}}{{{{\cos }^2}x}}dx} \\I = \int {\frac{2}{{{{\cos }^2}x}}dx} - \int {\frac{{3\sin x}}{{{{\cos }^2}x}}dx} \\I = 2\int {{{\sec }^2}} xdx - 3\int {\tan x\sec xdx} \\I = 2\tan x - 3\sec x + c\end{array}$

Choose the correct answer in Exercise 21 and 22

Question 21:

The anti derivative of $\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)$ equals
${(A)\frac{1}{3}{x^{\frac{1}{3}}} + 2{x^{\frac{1}{2}}} + c}$
${(B)\frac{2}{3}{x^{\frac{2}{3}}} + \frac{1}{2}{x^2} + c}$
${(C)\frac{2}{3}{x^{\frac{2}{3}}} + 2{x^{\frac{1}{2}}} + c}$

${(D)\frac{3}{2}{x^{\frac{3}{2}}} + \frac{1}{2}{x^{\frac{1}{2}}} + c}$
Solution: Antiderivative of is a function whose derivative will be (√x+ 1/√x
${\frac{d}{{dx}}\left( {{x^{\frac{3}{2}}}} \right) = \frac{3}{2}{x^{\frac{1}{2}}} = \frac{3}{2}\sqrt x }$ ${ \Rightarrow \sqrt x = \frac{2}{3}\frac{d}{{dx}}\left( {{x^{\frac{3}{2}}}} \right)}$ ${\sqrt x = \frac{d}{{dx}}\left( {\frac{2}{3}{x^{\frac{3}{2}}}} \right)}$ Now
${\frac{d}{{dx}}\sqrt x = \frac{1}{{2\sqrt x }}}$ ${ \Rightarrow \frac{1}{{\sqrt x }} = \frac{d}{{dx}}\left( {2\sqrt x } \right)}$ ∴ Antiderivative of √x + 1/√x is $\frac{2}{3}{x^{\frac{3}{2}}} + 2\sqrt x + c$

Question (22):

if
$\frac{d}{{dx}}f\left( x \right) = 4{x^3} - \frac{3}{{{x^4}}}$ Such that f(2)=0. Then f(x) is $\begin{array}{l}\left( A \right){x^4} + \frac{1}{{{x^3}}} - \frac{{129}}{8}\\\left( B \right){x^3} + \frac{1}{{{x^4}}} + \frac{{129}}{8}\\\left( C \right){x^3} + \frac{1}{{{x^3}}} + \frac{{129}}{8}\\\left( D \right){x^3} + \frac{1}{{{x^4}}} - \frac{{129}}{8}\end{array}$ Solution:
Derivative of integration are inverse function of each other
$\begin{array}{l}f\left( x \right) = \int {\frac{d}{{dx}}} f\left( x \right)dx\\f\left( x \right) = \int {\left( {4{x^3} - \frac{3}{{{x^4}}}} \right)dx} \\f\left( x \right) = 4\int {{x^3}dx - 3\int {{x^{ - 4}}dx} } \\f\left( x \right) = 4\frac{{{x^4}}}{4} - 3\frac{{{x^{ - 3}}}}{{ - 3}} + c\\f\left( x \right) = {x^4} + \frac{1}{{{x^3}}} + c\end{array}$
To get value of "c" we will use f(2) = 0
$\begin{array}{l}{\left( 2 \right)^4} + \frac{1}{{{{\left( 2 \right)}^3}}} + c = 0\\16 + \frac{1}{8} + c = 0\\c = \frac{{ - 129}}{8}\\f\left( x \right) = {x^4} + \frac{1}{{{x^3}}} - \frac{{129}}{8}\end{array}$ ∴ Option (A) is correct