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

$\begin{array}{l} \text { 1. } \displaystyle\int k \ \mathrm{d}x=k x+c \\\\ \text { 2. } \displaystyle\int f^{\prime}(x) \ \mathrm{d}x=f(x)+c \\\\ \text { 3. } \displaystyle\int x^{n} \ \mathrm{d}x=\dfrac{x^{n+1}}{n+1}+c \\\\ \text { 4. } \displaystyle\int k f(x) \ \mathrm{d}x=k \displaystyle\int f(x) \ \mathrm{d}x\\\\ \text { 5. } \displaystyle\int\left[f(x) \pm g(x)\right] \ \mathrm{d}x = \displaystyle\int f(x) \ \mathrm{d}x \pm \displaystyle\int g(x) \ \mathrm{d}x \end{array}$

1. Integrate each of the following with respect to $x$.
   (a) $x^3$
   (b) $3\sqrt{x}$
   (c) $\dfrac{2}{x^2}$
   (d) $\dfrac{1}{2\sqrt{x}}$
   (e) $(3x+5)\ \mathrm{d}x$





$\begin{aligned} \text{(a)}\ \displaystyle\int x^3\ \mathrm{d}x &=\dfrac{x^{3+1}}{3+1}+C\\\\ &=\dfrac{x^4}{4}+C \end{aligned}$ $\begin{aligned} \text{(b)}\ \displaystyle\int 3\sqrt{x} \ \mathrm{d}x &=3\cdot \displaystyle\int \sqrt{x}\ \mathrm{d}x\\\\ &=3\cdot \displaystyle\int x^{\frac{1}{2}}\ \mathrm{d}x\\\\ &=3\cdot \dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\\\\ &=2x^{\frac{3}{2}}+C \end{aligned}$ $\begin{aligned} \text{(c)}\ \displaystyle\int \dfrac{2}{x^2}\ \mathrm{d}x &=2\cdot \displaystyle\int x^{-2}\ \mathrm{d}x\\\\ &=2\cdot \dfrac{x^{-2+1}}{-2+1} + C\\\\ &=-\dfrac{2}{x}+C \end{aligned}$ $\begin{aligned} \text{(d)}\ \displaystyle\int \dfrac{2}{x^2}\ \mathrm{d}x &=\dfrac{1}{2}\cdot \displaystyle\int \dfrac{1}{x^{\frac{1}{2}}}\ \mathrm{d}x\\\\ &=\dfrac{1}{2}\cdot \displaystyle\int x^{-\frac{1}{2}}\ \mathrm{d}x\\\\ &=\dfrac{1}{2}\cdot \dfrac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C\\\\ &=\sqrt{x}+C \end{aligned}$ $\begin{aligned} \text{(e)}\ \displaystyle\int (3x+5)\ \mathrm{d}x &= \displaystyle\int (3x)\ \mathrm{d}x+ \displaystyle\int (5)\ \mathrm{d}x\\\\ &= \frac{3x^2}{2}+5x+C \end{aligned}$



2. Find each of the following indefinite integrals.
   
   (a) $ \displaystyle\int (3x − 1)(x + 2)\ \mathrm{d}x$
   
   (b) $ \displaystyle\int\left(3 x^{3}-4 \sqrt{x}+3\right) \ \mathrm{d}x$
   
   (c) $ \displaystyle\int\left(6 x^{2}-\dfrac{4}{x^{2}}\right) \ \mathrm{d}x$
   
   (d) $ \displaystyle\int\left(5-\dfrac{1}{\sqrt{x}}+\dfrac{1}{x^{3}}\right) \ \mathrm{d}x$
   
   (e) $ \displaystyle\int \dfrac{x^{4}+5 x}{2 x^{3}} \ \mathrm{d}x$





$\begin{aligned} \text{(a)}\ \quad\quad &\displaystyle\int \left(3x-1\right)\left(x+2\right)\ \mathrm{d}x\\\\ =& \displaystyle\int (3x^2+5x-2)\ \mathrm{d}x\\\\ =& \displaystyle\int 3x^2\ \mathrm{d}x+ \displaystyle\int 5x\ \mathrm{d}x- \displaystyle\int 2\ \mathrm{d}x\\\\ =& x^3+\frac{5x^2}{2}-2x+C\\\\ \end{aligned}$ $\begin{aligned} \text{(b)}\quad\quad& \displaystyle\int\left(3 x^{3}-4 \sqrt{x}+3\right) \mathrm{d} x \\\\ =& \displaystyle\int 3 x^{3} \mathrm{~d} x-\displaystyle\int 4 \sqrt{x} \mathrm{~d} x+\displaystyle\int 3 \mathrm{~d} x \\\\ =& \displaystyle\int 3 x^{3} \ \mathrm{d}x-\displaystyle\int 4 x^{\frac{1}{2}} \ \mathrm{d}x+\displaystyle\int 3 \ \mathrm{d}x \\\\ =& \dfrac{3 x^{4}}{4}-\dfrac{4 x^{\frac{3}{2}}}{\frac{3}{2}}+3 x+C \\\\ =& \dfrac{3}{4} x^{4}-\dfrac{8}{3} x^{\frac{3}{2}}+3 x+C \end{aligned}$ $\begin{aligned} \text{(c)}\quad\quad & \displaystyle\int\left(6 x^{2}-\dfrac{4}{x^{2}}\right) \ \mathrm{d}x \\\\ =& \displaystyle\int 6 x^{2} \ \mathrm{d}x-\displaystyle\int \dfrac{4}{x^{2}} \ \mathrm{d}x \\\\ =& \displaystyle\int 6 x^{2} \ \mathrm{d}x-\displaystyle\int 4 x^{-2} \ \mathrm{d}x \\\\ =& \dfrac{6 x^{3}}{3}-\dfrac{4 x^{-1}}{-1}+C \\\\ =& 2 x^{3}+\dfrac{4}{x} \end{aligned}$ $\begin{aligned} \text{ (d) }\quad\quad & \displaystyle\int\left(5-\dfrac{1}{\sqrt{x}}+\dfrac{1}{x^{3}}\right) \mathrm{~d} x \\\\ =& \displaystyle\int 5 \mathrm{~d} x-\displaystyle\int x^{-\frac{1}{2}} \mathrm{~d} x+\displaystyle\int x^{-3} \mathrm{~d} x \\\\ =& 5 x-\dfrac{x^{\frac{1}{2}}}{\frac{1}{2}}+\frac{x^{-2}}{2}+C \\\\ =& 5 x-2 \sqrt{x}+\dfrac{1}{2 x^{2}}+C \end{aligned}$ $\begin{aligned} \text{(e)}\quad\quad& \displaystyle\int \dfrac{x^{4}+5 x}{2 x^{3}} \mathrm{~d} x \\\\ =& \displaystyle\int \dfrac{x^{4}}{2 x^{3}} \mathrm{~d} x+\displaystyle\int \dfrac{5 x}{2 x^{3}} \mathrm{~d} x \\\\ =& \displaystyle\int \dfrac{1}{2} x \mathrm{~d} x+\displaystyle\int \dfrac{5}{2} x^{-2} \mathrm{~d} x \\\\ =& \dfrac{1}{2} \cdot \dfrac{x^{2}}{2}+\dfrac{5}{2} \dfrac{x^{-1}}{-1}+C \\\\ =& \dfrac{1}{4} x^{2}-\dfrac{5}{2 x}+C \end{aligned}$



3. Find each of the following indefinite integrals.
   
   (a) $\displaystyle\int \dfrac{3 x}{2 \sqrt[5]{x^{2}}} \mathrm{~d} x$
   
   (b) $\displaystyle\int \dfrac{(3 x-1)^{2}}{5 x^{4}} \mathrm{~d} x$
   
   (c) $\displaystyle\int \dfrac{3 x^{7}+x^{2}}{2 \sqrt[3]{x}} \mathrm{~d} x$
   
   (d) $\displaystyle\int(x-3 \sqrt{x})^{2} \mathrm{~d} x$
   
   (e) $\displaystyle\int(1+\sqrt[4]{x})(1-\sqrt[4]{x}) \mathrm{d} x$
   
   (f) $\displaystyle\int\left(\sqrt[3]{x}+\dfrac{2}{\sqrt[3]{x}}\right)^{2} \mathrm{~d} x$





$\begin{aligned} \text { (a) } \quad\quad&\displaystyle\int \frac{3 x}{2 \sqrt[5]{x^{2}}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3 x}{2 x^{\frac{2}{5}}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3}{2} \cdot x^{-\frac{3}{2}} \mathrm{~d} x \\\\ =&\dfrac{3}{2} \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}+C \\\\ =&-\dfrac{3}{\sqrt{x}}+C \end{aligned}$ $\begin{aligned} \text { (b) } \quad\quad&\displaystyle\int \frac{(3 x-1)^{2}}{5 x^{4}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{9 x^{2}-6 x+1}{5 x^{4}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{9}{5} x^{-2} \mathrm{~d} x-\displaystyle\int \dfrac{6}{5} x^{-3} \mathrm{~d} x+\displaystyle\int \dfrac{1}{5} x^{-4} \mathrm{~d} x \\\\ =&\dfrac{9}{5} \dfrac{x^{-1}}{-1}-\dfrac{6}{5} \dfrac{x^{-2}}{-2}+\dfrac{1}{5} \dfrac{x^{-3}}{-3}+C \\\\ =&-\dfrac{9}{5 x}+\dfrac{6}{5 x^{2}}-\dfrac{1}{15 x^{3}}+C \end{aligned}$ $\begin{aligned} \text { (c) }\quad\quad& \displaystyle\int \dfrac{3 x^{7}+x^{2}}{2 \sqrt[3]{x}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3 x^{7}}{2 \sqrt[3]{x}} \mathrm{~d} x+\displaystyle\int \dfrac{x^{2}}{2 \sqrt[3]{x}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3}{2} \dfrac{x^{7}}{x^{\frac{1}{3}}} \mathrm{~d} x+\displaystyle\int \dfrac{1}{2} \dfrac{x^{2}}{x^{\frac{1}{3}}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3}{2} x^{\frac{20}{3}} \mathrm{~d} x+\displaystyle\int \dfrac{1}{2} x^{\frac{5}{3}} \mathrm{~d} x\\\\ =&\dfrac{3}{2} \dfrac{x^{\frac{23}{3}}}{\frac{23}{3}}+\dfrac{1}{2} \dfrac{x^{\frac{8}{3}}}{\frac{8}{3}}+C \\ =&\dfrac{9}{46} x^{\frac{23}{3}}+\dfrac{3}{16} x^{\frac{8}{3}}+C \end{aligned}$ $\begin{aligned} \text { (d) } \quad\quad&\displaystyle\int(x-3 \sqrt{x})^{2} \mathrm{~d} x \\\\ =&\displaystyle\int\left(x^{2}-6 x \sqrt{x}+9 x\right) \mathrm{~d} x \\\\ =&\displaystyle\int x^{2} \mathrm{~d} x-\displaystyle\int 6 x^{\frac{3}{2}} \mathrm{~d} x+9 x \mathrm{~d} x \\\\ =&\dfrac{x^{2}}{3}-6 \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}+9 \dfrac{x^{2}}{2}+C \\\\ =&\dfrac{1}{3} x^{2}-4 x^{\frac{3}{2}}+\dfrac{9}{2} x^{2}+C \end{aligned}$ $\begin{aligned} \text { (e) } \quad\quad&\displaystyle\int(1+\sqrt[4]{x})(1-\sqrt[4]{x}) \mathrm{d} x \\\\ =&\displaystyle\int\left(1-(\sqrt[4]{x})^{2}\right) \mathrm{~d} x \\\\ =&\displaystyle\int\left(1-x^{\frac{1}{2}}\right) \mathrm{~d} x \\\\ =&\displaystyle\int 1 \mathrm{~d} x-\displaystyle\int x^{1 / 2} \mathrm{~d} x \\\\ =&x-\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}+C \\\\ =&x-\dfrac{2}{3} x^{\frac{3}{2}}+C \end{aligned}$ $\begin{aligned} \text { (f) } \quad\quad&\displaystyle\int\left(\sqrt[3]{x}+\frac{2}{\sqrt[3]{x}}\right)^{2} \mathrm{~d} x \\\\ =& \displaystyle\int\left(x^{\frac{1}{3}}+\dfrac{2}{x^{\frac{1}{3}}}\right)^{2} \mathrm{~d} x \\\\ =& \displaystyle\int\left(x^{\frac{2}{3}}+4+\dfrac{4}{x^{\frac{2}{3}}}\right) \mathrm{~d} x \\\\ =& \displaystyle\int x^{\frac{3}{2}} \mathrm{~d} x+\displaystyle\int 4 \mathrm{~d} x+\displaystyle\int 4 x^{-\frac{2}{3}} \mathrm{~d} x \\\\ =& \dfrac{x^{\frac{5}{2}}}{\frac{5}{2}}+4 x+4 \cdot \dfrac{x^{\frac{1}{3}}}{\frac{1}{3}}+C \\\\ =& \dfrac{2}{5} x^{\frac{5}{2}}+4 x+12 x^{\frac{1}{3}}+C \end{aligned}$



4. The rate of change of A with respect to r is given by $\dfrac{dA}{dr}= 4r+7$. If $A = 12$ when $r = 1$,find $A$ in terms of $r$.





$\begin{aligned} \dfrac{\mathrm{~d} A}{\mathrm{~d} r}=&4 r+7 \\\\ A=&\displaystyle\int \mathrm{~d} A \\\\ =&\displaystyle\int \frac{\mathrm{~d} A}{\mathrm{~d} r} \cdot \mathrm{~d} r \\\\ =&\displaystyle\int(4 r+7) \mathrm{~d} r\\\\ =&\displaystyle\int 4 r \cdot \mathrm{~d} r+\displaystyle\int 7 \mathrm{~d} r \\\\ =&2 r^{2}+7 r+C \\\\ \text { When }\ r=1,\ & A=12 \\\\ \therefore\ 12=&2+7+C \\\\ \therefore\ C=&3 . \\\\ \therefore\ A=&2 r^{2}+7 r+3 \end{aligned}$



5. 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.





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 $y=\dfrac{2}{3} x^{3}+\dfrac{7}{2} x^{2}$



6. 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.





$\begin{aligned} \dfrac{d y}{\mathrm{~d}x} &=k \sqrt[3]{x} \\\\ y &=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{d y}{\mathrm{~d}x} \cdot \mathrm{~d}x \\\\ &=\displaystyle\int k \sqrt[3]{x} \mathrm{~d}x \\\\ &=k \displaystyle\int x^{\frac{1}{3}} \mathrm{~d}x \\\\ &=\dfrac{3 k}{4} x^{\frac{4}{3}}+C\\\\ \end{aligned}$ 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 $(2)$, $\begin{array}{l} k=\dfrac{16}{15}, c=\dfrac{16}{5} \\\\ \therefore\ y=\dfrac{4}{5} x^{\frac{4}{3}}+\dfrac{16}{5} \end{array}$



7. 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.





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



8. 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.
   





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



9. 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.





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