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Indefinite Integration

Antiderivative ဟာ Grade 12 သင်ရိုးတွင် ပြဌာန်းလာမည့် သင်ခန်းစာတစ်ခုဖြစ်ပါတယ်။ သင်ခန်းစာရှင်းလင်းချက်များကို ဒီနေရာမှာ ရေခဲ့ဖူးပါတယ်။ အဆိုပါ post နှင့် ယှဉ်တွဲလေ့လာပြီး အောက်ပါ လေ့ကျင့်ခန်းများကို လေ့လာ လေ့ကျင့်ကြည့်နိုင်ပါတယ်။ စဉ်ဆက်မပြတ် လေ့လာသင်ယူနိုင်ကြပါစေ။


Rules of Integration

 1. k dx=kx+c 2. f(x) dx=f(x)+c 3. xn dx=xn+1n+1+c 4. kf(x) dx=kf(x) dx 5. [f(x)±g(x)] dx=f(x) dx±g(x) dx
  1. Integrate each of the following with respect to x.
    (a) x3
    (b) 3x
    (c) 2x2
    (d) 12x
    (e) (3x+5) dx


  2. (a) x3 dx=x3+13+1+C=x44+C

    (b) 3x dx=3x dx=3x12 dx=3x12+112+1+C=2x32+C

    (c) 2x2 dx=2x2 dx=2x2+12+1+C=2x+C

    (d) 2x2 dx=121x12 dx=12x12 dx=12x12+112+1+C=x+C

    (e) (3x+5) dx=(3x) dx+(5) dx=3x22+5x+C

  3. Find each of the following indefinite integrals.

    (a) (3x1)(x+2) dx

    (b) (3x34x+3) dx

    (c) (6x24x2) dx

    (d) (51x+1x3) dx

    (e) x4+5x2x3 dx


  4. (a) (3x1)(x+2) dx=(3x2+5x2) dx=3x2 dx+5x dx2 dx=x3+5x222x+C

    (b)(3x34x+3)dx=3x3 dx4x dx+3 dx=3x3 dx4x12 dx+3 dx=3x444x3232+3x+C=34x483x32+3x+C

    (c)(6x24x2) dx=6x2 dx4x2 dx=6x2 dx4x2 dx=6x334x11+C=2x3+4x

     (d) (51x+1x3) dx=5 dxx12 dx+x3 dx=5xx1212+x22+C=5x2x+12x2+C

    (e)x4+5x2x3 dx=x42x3 dx+5x2x3 dx=12x dx+52x2 dx=12x22+52x11+C=14x252x+C

  5. Find each of the following indefinite integrals.

    (a) 3x25x2 dx

    (b) (3x1)25x4 dx

    (c) 3x7+x223x dx

    (d) (x3x)2 dx

    (e) (1+4x)(14x)dx

    (f) (3x+23x)2 dx


  6.  (a) 3x25x2 dx=3x2x25 dx=32x32 dx=32x1212+C=3x+C

     (b) (3x1)25x4 dx=9x26x+15x4 dx=95x2 dx65x3 dx+15x4 dx=95x1165x22+15x33+C=95x+65x2115x3+C

     (c) 3x7+x223x dx=3x723x dx+x223x dx=32x7x13 dx+12x2x13 dx=32x203 dx+12x53 dx=32x233233+12x8383+C=946x233+316x83+C

     (d) (x3x)2 dx=(x26xx+9x) dx=x2 dx6x32 dx+9x dx=x236x3232+9x22+C=13x24x32+92x2+C

     (e) (1+4x)(14x)dx=(1(4x)2) dx=(1x12) dx=1 dxx1/2 dx=xx3232+C=x23x32+C

     (f) (3x+23x)2 dx=(x13+2x13)2 dx=(x23+4+4x23) dx=x32 dx+4 dx+4x23 dx=x5252+4x+4x1313+C=25x52+4x+12x13+C

  7. The rate of change of A with respect to r is given by dAdr=4r+7. If A=12 when r=1,find A in terms of r.


  8.  dA dr=4r+7A= dA= dA dr dr=(4r+7) dr=4r dr+7 dr=2r2+7r+C When  r=1, A=12

  9. Given that the gradient of a curve is 2x^2 + 7x and that the curve passes through the origin, determine the equation of the curve.


  10. Let the given curve be y.
    \begin{aligned} &\quad\text { Gradient of the curve } \\\\ &=\dfrac{\mathrm{~d}y}{\mathrm{~d}x}\\\\ &=2 x^{2}+7 x \\\\ &y=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{\mathrm{~d}y}{\mathrm{~d}x} \mathrm{~d}x \\\\ &=\displaystyle\int\left(2 x^{2}+7 x\right) \mathrm{~d}x \\\\ &=\dfrac{2}{3} x^{3}+\dfrac{7}{2} x^{2}+C \end{aligned}
    through the origin (0,0),
    \begin{aligned} 0 &=\dfrac{3}{2}(0)^{3}+\dfrac{7}{2}(0)^{2}+C \\\\ \therefore C &=0 \end{aligned}
    Hence, the equation the curve is y=\dfrac{2}{3} x^{3}+\dfrac{7}{2} x^{2}

  11. A curve is such that \dfrac{dy}{dx}=k\sqrt[3]{x} , where k is a constant and that it passes through the points (1, 4) and (8, 16). Find the equation of the curve.


  12. \begin{aligned} \dfrac{d y}{\mathrm{~d}x} &=k \sqrt[3]{x} \\\\ y &=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{d y}{\mathrm{~d}x} \cdot \mathrm{~d}x \\\\ &=\displaystyle\int k \sqrt[3]{x} \mathrm{~d}x \\\\ &=k \displaystyle\int x^{\frac{1}{3}} \mathrm{~d}x \\\\ &=\dfrac{3 k}{4} x^{\frac{4}{3}}+C\\\\ \end{aligned}
    Since the curve passes through the point (1,4) and (8,16)
    \begin{array}{l} 4=\dfrac{3 k}{4}+c\\\\ \therefore\ 3 k+4 C=16 \ldots(1)\\\\ 16=\dfrac{3 k}{4}(8)^{\frac{4}{3}}+C\\\\ 12 k+C=16\ldots(2) \end{array}
    Solving equatione (1) and (2),
    \begin{array}{l} k=\dfrac{16}{15}, c=\dfrac{16}{5} \\\\ \therefore\ y=\dfrac{4}{5} x^{\frac{4}{3}}+\dfrac{16}{5} \end{array}

  13. The gradient of a curve at the point (x, y) on the curve is given by \dfrac{x^{2}-4}{x^{2}}. Given that the curve passes through the point (2,7), find the equation of the curve.


  14. \begin{aligned} \dfrac{d y}{\mathrm{~d}x} &=\dfrac{x^{2}-4}{x^{2}} \\\\ y &=\displaystyle \int \mathrm{~d}y \\\\ &=\displaystyle \int \dfrac{d y}{\mathrm{~d}x} \cdot \mathrm{~d}x \\\\ &=\displaystyle \int \dfrac{x^{2}-4}{x^{2}} \mathrm{~d}x \\\\ &=\displaystyle \int\left(1-\dfrac{4}{x^{2}}\right) \mathrm{~d}x \\\\ &=\displaystyle \int 1 \mathrm{~d}x-\displaystyle \int 4 x^{-2} \mathrm{~d}x \\\\ &=x+\dfrac{4}{x}+C \end{aligned}
    Since the curve passes through the point (2,7), 7=2+\dfrac{4}{2}+C
    C=3
    \therefore\ y=x+\dfrac{4}{x}+3

  15. A curve with \dfrac{dy}{dx}=k x+3, where k is a constant, passes through the point P(3,19). Given that the gradient of the normal to the curve at the point P is -\dfrac{1}{15}, find
    (i) the value of k,
    (ii) the equation of the curve,
    (iii) the coordinates of the turning point on the curve.


  16. \begin{aligned} \dfrac{\mathrm{~d}y}{\mathrm{~d}x}&=k x+3 \\\\ \text {gradient of normal at }&(3,19)=-\dfrac{1}{15} \\\\ \therefore-\left.\dfrac{1}{\dfrac{\mathrm{~d}y}{\mathrm{~d}x}}\right|_{(3,19)}&=-\dfrac{1}{15} \\\\ \left.\therefore \dfrac{1}{k x+3}\right|_{(3,19)}&=\dfrac{1}{15}\\\\ \therefore 3 k+3&=15 \\\\ k&=4 \\\\ \therefore \quad \dfrac{\mathrm{~d}y}{\mathrm{~d}x}&=4 x+3 \end{aligned}
    \begin{aligned} y&=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{\mathrm{~d}y}{\mathrm{~d}x} \cdot \mathrm{~d}x \\\\ &=\displaystyle\int(4 x+3) \mathrm{~d}x \\\\ &=2 x^{2}+3 x+C \end{aligned}
    Since the curve passe through the point (3,19)
    \begin{array}{l} 19 =2(3)^{2}+3(3)+C 19 =18+9+C \\\\ C =-8 \\\\ \therefore \quad\mathrm{~d}y=2 x^{2}+3 x-8\\\\ \text { At turning point, } \\\\ \dfrac{\mathrm{~d}y}{\mathrm{~d}x}=0 \\\\ 4 x+3=0\\\\ x=-\dfrac{3}{4} \\\\ y=2\left(-\dfrac{3}{4}\right)^{2}+3\left(-\dfrac{3}{4}\right)-8 \\\\ \quad =-\dfrac{73}{8} \\\\ \text { The turning point is } \left(-\dfrac{3}{4},-\dfrac{73}{8}\right) \end{array}

  17. The equation of a curve is such that \dfrac{dy}{dx}=\dfrac{1}{(x-3)^{2}}+x . It is given that the curve passes through the point (2,7). Find the equation of the curve.


  18. \begin{aligned} \dfrac{dy}{dx} &=\dfrac{1}{(x-3)^{2}}+x \\\\ y &=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{dy}{dx} \\\\ &=\displaystyle\int\left(\dfrac{1}{(x-3)^{2}}+x\right) \mathrm{~d}x \\\\ &=\displaystyle\int \dfrac{1}{(x-3)^{2}} \mathrm{~d}x+\displaystyle\int x \mathrm{~d}x\\\\ \end{aligned}
    \begin{aligned} \text{Since}\ \dfrac{d}{\mathrm{~d}x}(x-3)&=1, \\\\ d(x-3)&=\mathrm{~d}x \end{aligned}

    \begin{aligned} y&=\displaystyle\int \dfrac{1}{(x-3)^{2}} d(x-3)+\displaystyle\int x \mathrm{~d}x \\\\ &=\displaystyle\int(x-3)^{-2} d(x-3)+\displaystyle\int x \mathrm{~d}x \\\\ &=-\dfrac{1}{x-3}+\dfrac{1}{2} x^{2}+c \end{aligned}
    Since (2,7) lies on the curve,
    \begin{array}{l} 7=-\dfrac{1}{2-3}+\dfrac{1}{2}(2)^{2}+C\\\\ C=4\\\\ \therefore\ y=\dfrac{1}{2} x^{2}-\dfrac{1}{x-3}+4 \end{array}

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