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NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.1 11–22 (part 2 of 27)

  1. Find the following integrals in Exercises $\displaystyle 6$ to $\displaystyle 20$:

    Exercise 11

    ∫x3+5x2−4x2dx\displaystyle \int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x

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    NCERT’s answer
    \(\displaystyle \frac{x^{2}}{2}+5 x+\frac{4}{x}+C\)
    Divide each term of the numerator by \(\displaystyle x^{2}\) first (\(\displaystyle x \neq 0\)): \[\frac{x^{3} + 5x^{2} - 4}{x^{2}} = x + 5 - \frac{4}{x^{2}} = x + 5 - 4x^{-2}.\] Integrate term by term with the power rule; note \(\displaystyle \int x^{-2}dx = \frac{x^{-1}}{-1} = -\frac{1}{x}\), so the last term changes sign: \[\int \left(x + 5 - 4x^{-2}\right) dx = \frac{x^{2}}{2} + 5x - 4\left(-\frac{1}{x}\right) + C.\] Hence \[\int \frac{x^{3} + 5x^{2} - 4}{x^{2}}\, dx = \frac{x^{2}}{2} + 5x + \frac{4}{x} + C, \qquad x \neq 0.\]

  2. Exercise 12

    ∫x3+3x+4xdx\displaystyle \int \frac{x^{3}+3 x+4}{\sqrt{x}} d x

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    NCERT’s answer
    \(\displaystyle \frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+C\)
    Write \(\displaystyle \sqrt{x} = x^{1/2}\) and divide each term of the numerator by it, subtracting exponents (\(\displaystyle x > 0\)): \[\frac{x^{3} + 3x + 4}{\sqrt{x}} = x^{5/2} + 3x^{1/2} + 4x^{-1/2}.\] Apply the power rule \(\displaystyle \int x^{n}dx = \frac{x^{n+1}}{n+1}\) to each term: \[\int x^{5/2}dx = \frac{x^{7/2}}{7/2} = \frac{2}{7}x^{7/2}, \quad 3\int x^{1/2}dx = 3\cdot\frac{x^{3/2}}{3/2} = 2x^{3/2}, \quad 4\int x^{-1/2}dx = 4\cdot\frac{x^{1/2}}{1/2} = 8x^{1/2}.\] Adding, \[\int \frac{x^{3} + 3x + 4}{\sqrt{x}}\, dx = \frac{2}{7}x^{7/2} + 2x^{3/2} + 8\sqrt{x} + C, \qquad x > 0.\]

  3. Exercise 13

    ∫x3−x2+x−1x−1dx\displaystyle \int \frac{x^{3}-x^{2}+x-1}{x-1} d x

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    NCERT’s answer
    \(\displaystyle \frac{x^{3}}{3}+x+\mathrm{C}\)
    Factorise the numerator by grouping so the denominator cancels: \[x^{3} - x^{2} + x - 1 = x^{2}(x - 1) + 1\,(x - 1) = (x - 1)\left(x^{2} + 1\right).\] Cancelling \(\displaystyle x - 1\) is legitimate only where \(\displaystyle x \neq 1\), which is exactly where the integrand is defined: \[\frac{x^{3} - x^{2} + x - 1}{x - 1} = x^{2} + 1, \qquad x \neq 1.\] Integrate by the power rule: \[\int \left(x^{2} + 1\right) dx = \frac{x^{3}}{3} + x + C.\] So \[\int \frac{x^{3} - x^{2} + x - 1}{x - 1}\, dx = \frac{x^{3}}{3} + x + C, \qquad x \neq 1.\]

  4. Exercise 14

    ∫(1−x)xdx\displaystyle \int(1-x) \sqrt{x} d x

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    NCERT’s answer
    \(\displaystyle \frac{2}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}+C\)
    Multiply out first, since there is no product rule for integrals: \[(1 - x)\sqrt{x} = x^{1/2} - x^{3/2}.\] Apply the power rule to each term: \[\int x^{1/2}dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}, \qquad \int x^{3/2}dx = \frac{x^{5/2}}{5/2} = \frac{2}{5}x^{5/2}.\] Hence \[\int (1 - x)\sqrt{x}\, dx = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C, \qquad x \geq 0.\]

  5. Exercise 15

    ∫x(3x2+2x+3)dx\displaystyle \int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x

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    NCERT’s answer
    \(\displaystyle \frac{6}{7} x^{\frac{7}{2}}+\frac{4}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+\mathrm{C}\)
    Distribute \(\displaystyle \sqrt{x} = x^{1/2}\) over the bracket, adding exponents: \[\sqrt{x}\left(3x^{2} + 2x + 3\right) = 3x^{5/2} + 2x^{3/2} + 3x^{1/2}.\] Integrate term by term with the power rule: \[3\cdot\frac{x^{7/2}}{7/2} = \frac{6}{7}x^{7/2}, \qquad 2\cdot\frac{x^{5/2}}{5/2} = \frac{4}{5}x^{5/2}, \qquad 3\cdot\frac{x^{3/2}}{3/2} = 2x^{3/2}.\] So \[\int \sqrt{x}\left(3x^{2} + 2x + 3\right) dx = \frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + 2x^{3/2} + C, \qquad x \geq 0.\]

  6. Exercise 16

    ∫(2x−3cos⁡x+ex)dx\displaystyle \int\left(2 x-3 \cos x+e^{x}\right) d x

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    NCERT’s answer
    \(\displaystyle x^{2}-3 \sin x+e^{x}+\mathrm{C}\)
    Integrate term by term, using the power rule, \(\displaystyle \int \cos x\, dx = \sin x\) and \(\displaystyle \int e^{x}dx = e^{x}\): \[\int \left(2x - 3\cos x + e^{x}\right) dx = 2\cdot\frac{x^{2}}{2} - 3\sin x + e^{x} + C.\] Hence \[\int \left(2x - 3\cos x + e^{x}\right) dx = x^{2} - 3\sin x + e^{x} + C.\]
  7. Find the following integrals in Exercises $\displaystyle 6$ to $\displaystyle 20$:

    Exercise 17

    ∫(2x2−3sin⁡x+5x)dx\displaystyle \int\left(2 x^{2}-3 \sin x+5 \sqrt{x}\right) d x

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    NCERT’s answer
    \(\displaystyle \frac{2}{3} x^{3}+3 \cos x+\frac{10}{3} x^{\frac{3}{2}}+\mathrm{C}\)
    Integrate term by term. The one to watch is the sine: \(\displaystyle \int \sin x\, dx = -\cos x\), so the \(\displaystyle -3\sin x\) term integrates to \(\displaystyle +3\cos x\). \[2\int x^{2}dx = \frac{2}{3}x^{3}, \qquad -3\int \sin x\, dx = 3\cos x, \qquad 5\int x^{1/2}dx = 5\cdot\frac{x^{3/2}}{3/2} = \frac{10}{3}x^{3/2}.\] Hence \[\int \left(2x^{2} - 3\sin x + 5\sqrt{x}\right) dx = \frac{2}{3}x^{3} + 3\cos x + \frac{10}{3}x^{3/2} + C, \qquad x \geq 0.\]

  8. Exercise 18

    ∫sec⁡x(sec⁡x+tan⁡x)dx\displaystyle \int \sec x(\sec x+\tan x) d x

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    NCERT’s answer
    \(\displaystyle \tan x+\sec x+\mathrm{C}\)
    Expand the product so that each piece is a standard derivative: \[\sec x\,(\sec x + \tan x) = \sec^{2}x + \sec x \tan x.\] Now use the standard results \(\displaystyle \frac{d}{dx}(\tan x) = \sec^{2}x\) and \(\displaystyle \frac{d}{dx}(\sec x) = \sec x \tan x\), i.e. \[\int \sec^{2}x\, dx = \tan x, \qquad \int \sec x \tan x\, dx = \sec x.\] Therefore \[\int \sec x\,(\sec x + \tan x)\, dx = \tan x + \sec x + C,\] valid on any interval on which \(\displaystyle \cos x \neq 0\).
  9. Find the following integrals in Exercises $\displaystyle 6$ to $\displaystyle 20$:

    Exercise 19

    ∫sec⁡2xcosec⁡2xdx\displaystyle \int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x

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    NCERT’s answer
    \(\displaystyle \tan x-x+\mathrm{C}\)
    Convert to sines and cosines before integrating: \[\frac{\sec^{2}x}{\mathrm{cosec}^{2}x} = \frac{1/\cos^{2}x}{1/\sin^{2}x} = \frac{\sin^{2}x}{\cos^{2}x} = \tan^{2}x.\] There is no direct formula for \(\displaystyle \int \tan^{2}x\,dx\), so use the identity \(\displaystyle 1 + \tan^{2}x = \sec^{2}x\), i.e. \(\displaystyle \tan^{2}x = \sec^{2}x - 1\): \[\int \tan^{2}x\, dx = \int \left(\sec^{2}x - 1\right) dx = \tan x - x + C.\] Hence \[\int \frac{\sec^{2}x}{\mathrm{cosec}^{2}x}\, dx = \tan x - x + C,\] valid on any interval on which \(\displaystyle \cos x \neq 0\).

  10. Exercise 20

    ∫2−3sin⁡xcos⁡2xdx\displaystyle \int \frac{2-3 \sin x}{\cos^{2} x} d x

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    NCERT’s answer
    \(\displaystyle 2 \tan x-3 \sec x+\mathrm{C}\)
    Split the fraction over the common denominator \(\displaystyle \cos^{2}x\): \[\frac{2 - 3\sin x}{\cos^{2}x} = \frac{2}{\cos^{2}x} - 3\cdot\frac{\sin x}{\cos^{2}x} = 2\sec^{2}x - 3\sec x \tan x,\] since \(\displaystyle \dfrac{\sin x}{\cos^{2}x} = \dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x} = \sec x \tan x\). Using \(\displaystyle \int \sec^{2}x\,dx = \tan x\) and \(\displaystyle \int \sec x \tan x\,dx = \sec x\): \[\int \frac{2 - 3\sin x}{\cos^{2}x}\, dx = 2\tan x - 3\sec x + C,\] valid on any interval on which \(\displaystyle \cos x \neq 0\).
  11. Choose the correct answer in Exercises $\displaystyle 21$ and 22.

    Exercise 21

    The anti derivative of (x+1x)\displaystyle \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) equals (A) 13x13+2x12+C\displaystyle \frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+\mathrm{C} (B) 23x23+12x2+C\displaystyle \frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^{2}+\mathrm{C} (C) 23x32+2x12+C\displaystyle \frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+\mathrm{C} (D) 32x32+12x12+C\displaystyle \frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+\mathrm{C}

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    NCERT’s answer
    C
    An anti derivative of a function is any function whose derivative is that function, so compute the integral by the power rule \(\displaystyle \int x^{n}dx = \frac{x^{n+1}}{n+1}\) with \(\displaystyle n = \frac{1}{2}\) and \(\displaystyle n = -\frac{1}{2}\): \[\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx = \int \left(x^{1/2} + x^{-1/2}\right) dx = \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C = \frac{2}{3}x^{3/2} + 2x^{1/2} + C.\] Check by differentiating: \(\displaystyle \frac{d}{dx}\left(\frac{2}{3}x^{3/2} + 2\sqrt{x}\right) = x^{1/2} + x^{-1/2}\), as required. This matches option (C). \[\boxed{\text{(C)}\ \ \frac{2}{3}x^{3/2} + 2x^{1/2} + C}\]

  12. Exercise 22

    If ddxf(x)=4x3−3x4\displaystyle \frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}} such that f(2)=0\displaystyle f(2)=0. Then f(x)\displaystyle f(x) is (A) x4+1x3−1298\displaystyle x^{4}+\frac{1}{x^{3}}-\frac{129}{8} (B) x3+1x4+1298\displaystyle x^{3}+\frac{1}{x^{4}}+\frac{129}{8} (C) x4+1x3+1298\displaystyle x^{4}+\frac{1}{x^{3}}+\frac{129}{8} (D) x3+1x4−1298\displaystyle x^{3}+\frac{1}{x^{4}}-\frac{129}{8}

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    NCERT’s answer
    A
    \(\displaystyle f\) is an anti derivative of the given derivative, so integrate, writing \(\displaystyle \frac{3}{x^{4}} = 3x^{-4}\): \[f(x) = \int \left(4x^{3} - 3x^{-4}\right) dx = 4\cdot\frac{x^{4}}{4} - 3\cdot\frac{x^{-3}}{-3} + C = x^{4} + \frac{1}{x^{3}} + C.\] The sign is the trap: \(\displaystyle \int x^{-4}dx = \frac{x^{-3}}{-3}\), and the minus in front of \(\displaystyle 3x^{-4}\) cancels it, leaving \(\displaystyle +\frac{1}{x^{3}}\). Now use the condition \(\displaystyle f(2) = 0\) to fix \(\displaystyle C\): \[2^{4} + \frac{1}{2^{3}} + C = 0 \quad \Longrightarrow \quad 16 + \frac{1}{8} + C = 0 \quad \Longrightarrow \quad C = -\frac{129}{8}.\] Therefore \[f(x) = x^{4} + \frac{1}{x^{3}} - \frac{129}{8},\] which is option (A).