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NCERT Solutions for Class 12 Maths Chapter 7: Integrals

In this page we have NCERT Solutions for Class 12 Maths Chapter 7: Integrals for EXERCISE 7.1 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Find an anti derivative (or integral) of the following functions by the method of inspection

Question 1: sin 2x.
Solution
The anti derivative of sin 2x is a function of x whose derivative is sin 2x.It is known that,
$\frac{d}{dx} (cos\; 2x) = – 2 sin\; 2x \\ sin\; 2x = – \frac{1}{2} \frac{d}{dx} (cos\; 2x) \\ sin\; 2x = \frac{d}{dx} (- \frac{1}{2} cos\; 2x) \\ Therefore,\; the\; anti – derivative\; of\; sin\; 2x \;is\; (- \frac{1}{2} cos\; 2x)$

Question 2:cos 3x.
Solution
The anti derivative of cos 3x is a function of x whose derivative is cos 3x.It is known that,
$\frac{d}{dx} (sin\; 3x) = 3 cos\; 3x \\ cos\; 3x = \frac{1}{3} \frac{d}{dx} (sin\; 3x) \\ cos\; 3x = \frac{d}{dx} (\frac{1}{3} (sin\; 3x)) \\ Therefore,\; the\; anti – derivative\; of\; cos\; 3x \;is\; (\frac{1}{3} sin\; 3x)$

Question 3:e^2x.
Solution
The anti derivative of e^2x is a function of x whose derivative is e^2x.It is known that,
$\frac{d}{dx} (e ^{2x}) = 2 e ^{2x} \\ e ^{2x} = \frac{1}{2} \frac{d}{dx} (e^{2x}) \\ e ^{2x} = \frac{d}{dx} (\frac{1}{2} e^{2x}) \\ Therefore,\; the\; anti – derivative\; of\; e ^{2x} is \frac{1}{2} e^{2x}$

Question 4: (ax + b) ².
Solution
The anti derivative of (ax + b) ² is a function of x whose derivative is (ax + b) ².It is known that,
$\frac{d}{dx} (ax + b)^{3} = 3a (ax + b) ^{2}\\ (ax + b) ^{2} = \frac{1}{3a} \frac{d}{dx} (ax + b) ^{3} \\ (ax + b) ^{2} = \frac{d}{dx} (\frac{1}{3a} (ax + b) ^{3}) \\ Therefore,\; the\; anti – derivative\; of\; (ax + b) ^{2} is \frac{1}{3a} (ax + b) ^{3}$

Question 5: sin 2x – 4 e^3x
Solution
The anti derivative of sin 2x – 4 e^3x is a function of x whose derivative is sin 2x – 4 e^3x.It is known that,
$\frac{d}{dx} (- \frac{1}{2} cos\; 2x – \frac{4}{3} e^{3x}) = sin 2x – 4 e^{3x} \\ Therefore,\; the\; anti – derivative\; of\; sin\; 2x – 4 e^{3x} \;is\; (- \frac{1}{2} cos\; 2x – \frac{4}{3} e^{3x})$

Find the following integrals in Exercises 6 to 20:

Question 6:$\int (4 e^{3x} + 1) dx$
Solution
$\int (4 e^{3x} + 1) dx \\ =4 \int e^{3x} du + \int 1 dx \\ =4 (\frac{e^{3x}}{3}) + x + C \\ =(4/3) e^{3x} + x + C \\ Where\; c\; is\; the\; constant.$

Question 7: $\int x^{2} (1 – \frac{1}{x^{2}}) dx$
Solution
$\int x^{2} (1 – \frac{1}{x^{2}}) dx \\ =\int (x^{2} – 1) dx \\ =\frac{x^{3}}{3} – x + C \\ Where\; c\; is\; the\; constant$

Question 8:$\int (a x^{2} + b x + c) du$
Solution
$\int (a x^{2} + b x + c) dx \\ =a \int (x^{2}) dx + b \int x dx + c \int 1 dx \\ =a (\frac{x^{3}}{3}) + b (\frac{x^{2}}{2}) + cx + D \\ \\ Where\; D\; is\; the\; constant$

Question 9:$\int (2 x^{2} + e^{x}) dx$
Solution
$\int (2 x^{2} + e^{x}) dx \\ =2 \int (x^{2}) dx + \int e^{x} dx \\ =2 (\frac{x^{3}}{3}) + e^{x} + C \\ \\ Where\; C\; is\; the\; constant$

Question 10: $\int (\sqrt{x} - \frac{1}{\sqrt{x}}) ^{2} dx$
Solution
$\int(\sqrt{x} -\frac{1}{\sqrt{x}}) ^{2} \\ =\int (x + \frac{1}{x} – 2) dx \\ = \int x dx + \int \frac{1}{x} dx – 2 \int 1 dx \\ =\frac{x^{2}}{2} + log \left | x \right | – 2 x + C \\ Where\; C\; is\; the\; constant$

Question 11: $\int \frac{x^{3} + 5 x^{2} - 4}{x ^{2}} dx$
Solution
$\int \frac{x^{3} + 5 x^{2} - 4}{x ^{2}} dx \\= \int x dx + 5 \int 1 dx - \int \frac{4}{x^{2}} dx \\ =\frac{x ^{2}}{2} + 5 x - \frac{4}{x} + C \\ Where\; C\; is\; the\; constant$

Question 12:$\int\frac{x^{3} + 3 x + 4}{\sqrt{x}}dx$
Solution
$\int \frac{x^{3} + 3 x + 4}{\sqrt{x}} dx \\ =\int (x ^{\frac{5}{2}} + 3 x ^{\frac{1}{2}} + 4 x^{- \frac{1}{2}}) \\ = \frac{x ^{\frac{7}{2}}}{\frac{7}{2}} + \frac{3 (x ^{\frac{3}{2}})}{\frac{3}{2}} + \frac{4 (x ^{\frac{1}{2}})}{\frac{1}{2}} + C \\ = \frac{2}{7} (x ^{\frac{7}{2}}) + 2 (x ^{\frac{3}{2}}) + 8 x ^{\frac{1}{2}} + C \\ = \frac{2}{7} (x ^{\frac{7}{2}}) + 2 (x ^{\frac{3}{2}}) + 8 \sqrt{x} + C \\ Where\; C\; is\; the\; constant$

Question 13: $\frac{x^{3} – x^{2} + x - 1}{x – 1}$
Solution
$\int \frac{x^{3} – x^{2} + x - 1}{x – 1} dx \\ \int \frac{(x^{2} + 1) ( x - 1)}{x – 1} dx \\ On\; divinding,\; we\; get\; \\ =\int (x^{2} + 1) dx \\ =\int x^{2} dx + \int 1 dx \\ =\frac{x^{3}}{3} + x + C Where\; C\; is\; the\; constant$

Question 14:$\int(1 – x) \sqrt{x}dx $
Solution
$\int (1 + x) \sqrt{x}\; dx \\ =\int (\sqrt{x} + x^{\frac{3}{2}}) dx \\ =\int x^{\frac{1}{2}} dx + \int x^{\frac{3}{2}} dx \\ =\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + C \\ =\frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} + C \\ Where\; C\; is\; the\; constant$

Question 15:$ \int\sqrt{x} (3x^{2} + 2x + 3) dx$
Solution
$\int \sqrt{x} (3x^{2} + 2x + 3) dx \\ =\int (3x ^{\frac{5}{2}} + 2x ^{\frac{3}{2}} + 3u ^{\frac{1}{2}}) dx \\ =3 \int x ^{\frac{5}{2}} dx + 2 \int x ^{\frac{3}{2}} dx + 3 \int x ^{\frac{1}{2}} dx \\ =3 (\frac{x ^{\frac{7}{2}}}{\frac{7}{2}}) + 2 (\frac{x ^{\frac{5}{2}}}{\frac{5}{2}}) + 3 (\frac{x ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ =\frac{6}{7} x ^{\frac{7}{2}} + \frac{4}{5} x ^{\frac{5}{2}} + 2 x ^{\frac{3}{2}} + C \\ Where\; C\; is\; the\; constant$

Question 16: $\int 2 x – 3 cos\; x + e ^{x} dx$
Solution
$\int (2 x – 3 cos\; x + e ^{x}) dx \\ =2 \int x dx – 3 \int cos\; x dx + \int e ^{x} dx \\ =2 \frac{x ^{2}}{2} – 3 (sin x) + e ^{x} + C \\ =x ^{2} – 2 sin\; x + e ^{x} + C Where\; C\; is\; the\; constant$

Question 17:$\int (2 x^{2} -3 sin x + 5 \sqrt{x})dx$
Solution
$\int (2 x^{2} - 3 sin x + 5 \sqrt{x})\; dx \\ =4 \int x^{2} dx - 3 \int sin x dx + 5 \int x^{\frac{1}{2}}dx \\ =\frac{2 x^{3}}{3} +3 (cos\; x) + 5 (\frac{x ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ =\frac{2}{3} x^{3} + 3 cos\; x + (10/3) x ^{\frac{3}{2}} + C \\ Where\; C\; is\; the\; constant$

Question 18: $\int sec\; x (tan\; x + sec\; x)$dx
Solution
$\int sec\; x (tan\; x + sec\; x) dx \\ =\int (sec\; x tan\;x + sec ^{2}\; x) dx \\ =sec\;x\; + tan\;x + C \\ Where\; C\; is\; the\; constant$

Question 19:$\int \frac{sec ^{2}\; x}{cosec ^{2}\; x}$dx
Solution
$\int \frac{sec ^{2}\; x}{cosec ^{2}\; x} dx \\ =\int \frac{\frac{1}{cos ^{2}\; x}}{\frac{1}{sin ^{2}\; x}} dx \\ =\int \frac{sin ^{2}\; x}{cos ^{2}\; x} dx \\ =\int (tan ^{2}\; x) dx \\ =\int (sec ^{2}\; x – 1) dx \\ =\int sec ^{2}\; x dx – \int 1 dx \\ =tan x – x + C \\ Where\; C\; is\; the\; constant$

Question 20: $\int \frac{3 – 2 sin\; x}{cos ^{2}\; x}dx$
Solution
$\int \frac{3 – 2 sin\; x}{cos ^{2}\; x} dx \\ =\int (\frac{3}{cos ^{2}\; x} – \frac{2 sin\; x}{cos ^{2}\; x}) dx \\ =3 \int sec ^{2}\; x; dx – 2 \int tan\; x\; sec\; x dx \\ =3 tan\; x – 2 sec\; x + C \\ Where\; C\; is\; the\; constant$

Question 21:
Which of the following below is an integral of $\sqrt{x} + \frac{1}{\sqrt{x}}dx$:
$(a) \frac{1}{3} x^{\frac{1}{3}} + 2 x^{\frac{1}{2}} + C \\ (b) \frac{2}{3} x^{\frac{2}{3}} + \frac{1}{2} x^{2} + C \\ (c) \frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + C \\ (d) \frac{3}{2} x^{\frac{3}{2}} + \frac{1}{2} x^{\frac{1}{2}} + C$ Solution
$\int \sqrt{x} + \frac{1}{\sqrt{x}} dx \\ =\int x ^{\frac{1}{2}} dx + \int x^{- \frac{1}{2}} dx \\ =\frac{x ^{\frac{3}{2}}}{\frac{3}{2}} + \frac{x ^{\frac{1}{2}}}{\frac{1}{2}} + C \\ =\frac{3}{2} x ^{\frac{3}{2}} + 2 x ^{\frac{1}{2}} \\ Option\; c\; is\; correct$

Question 22:
$if\; \frac{d}{dx} f (x) = 4 x^{3} – \frac{3}{x^{4}},\; in\; such\; a\; way\; that\; f (2) = 0,\; then\; f (x)\; is \\ (a) x^{4} + \frac{1}{x ^{3}} – \frac{129}{8} \\ (b) x^{3} + \frac{1}{x ^{4}} + \frac{129}{8} \\ (c) x^{4} + \frac{1}{x ^{3}} + \frac{129}{8} \\(d) x^{3} + \frac{1}{x ^{4}} – \frac{129}{8}$
Solution
Given,
$\frac{d}{dx} f (r) = 4 x^{3} – \frac{3}{x^{4}} \\ Integral\; of\; 4 x^{3} – \frac{3}{x^{4}} = f (x) \\ f (x) = \int 4 x^{3} – \frac{3}{x^{4}}\; dx \\ f (x) = 4 \int x^{3} dr – 3 \int (x ^{- 4}) dx \\ f (x) = 4 \frac{x ^{4}}{4} – 3 \frac{x ^{- 3}}{- 3} + K \\ f (x) = x ^{4} + \frac{1}{x ^{3}} + K $
And$ f (2) = 0 \\ f (2) = 2 ^{4} + \frac{1}{2 ^{3}} + K = 0 \\$
$16 + \frac{1}{8} + K = 0 \\ K = – \frac{129}{8} \\ f (x) = x ^{4} + \frac{1}{x ^{3}} – \frac{129}{8} \\ Option\; (a)\; is\; correct$