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) $(3x+5)\ \mathrm{d}x$
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 $y=\dfrac{2}{3} x^{3}+\dfrac{7}{2} x^{2}$
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 $(2)$, $\begin{array}{l} k=\dfrac{16}{15}, c=\dfrac{16}{5} \\\\ \therefore\ y=\dfrac{4}{5} x^{\frac{4}{3}}+\dfrac{16}{5} \end{array}$
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 $(2,7)$, $7=2+\dfrac{4}{2}+C$ $C=3$ $\therefore\ y=x+\dfrac{4}{x}+3$
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 $(3,19)$ $\begin{array}{l} 19 =2(3)^{2}+3(3)+C 19 =18+9+C \\\\ C =-8 \\\\ \therefore \quad\mathrm{~d}y=2 x^{2}+3 x-8\\\\ \text { At turning point, } \\\\ \dfrac{\mathrm{~d}y}{\mathrm{~d}x}=0 \\\\ 4 x+3=0\\\\ x=-\dfrac{3}{4} \\\\ y=2\left(-\dfrac{3}{4}\right)^{2}+3\left(-\dfrac{3}{4}\right)-8 \\\\ \quad =-\dfrac{73}{8} \\\\ \text { The turning point is } \left(-\dfrac{3}{4},-\dfrac{73}{8}\right) \end{array}$
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.
➋ Given that $ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}-4x-5$ and $ \displaystyle y=-4$ when $ \displaystyle x=-2,$ find the value of $ \displaystyle y$ when $ \displaystyle x=1.$
➍ Given that the curve which has gradient $ \displaystyle \frac{dy}{dx}=2x^2 + 7x$ at any point on the curve passes through the origin, determine the equation of the curve.
➎ The rate of change of $ \displaystyle A$ with respect to $ \displaystyle r$ is given by $ \displaystyle \frac{{dA}}{{dr}}=6r+1$. If $ \displaystyle A = 3$ when $ \displaystyle r = 1$, find $ \displaystyle A$ in terms of $ \displaystyle r$.
➏ If the curve which has gradient $ \displaystyle \frac{dy}{dx}=kx-1$ at any point on the curve cuts the $ \displaystyle x$-axis at $ \displaystyle x=-3$ and $ \displaystyle x=4$, determine the value of $ \displaystyle k$ and hence find the equation of the curve.
➑ Given that a ball moves along a grove with the velocity $ \displaystyle v=\frac{{ds}}{{dt}}=32t-2$ ft/s and at the time $ \displaystyle t=\frac{1}{2}\ \text{s}$ the ball is $ \displaystyle 4\ \text{ft}$ away from the starting point. Express the position of the ball as a function of time.
➓ Find the equation of the curve which cuts the $ \displaystyle y$-axis at $ \displaystyle (0, 4)$ and for which $ \displaystyle \frac{{dy}}{{dx}}={{(2x+1)}^{3}}$.
The reverse process of obtaining $ \displaystyle y$ from $ \displaystyle \frac{dy}{dx}$ is called integration. Therefore integration is the reverse of differentiation.
အပြန်အလှန်အားဖြင့် $ \displaystyle \frac{dy}{dx}$ ကနေ မူလ function function ($ \displaystyle y$) ပြန်ရအောင် လုပ်တဲ့ လုပ်ငန်းစဉ်ကိုတော့ anti-derivative (integration) လို့ ခေါ်ပါတယ်။
$ \displaystyle \frac{d}{{dx}}\left[ {kf\left( x \right)} \right]=k\ {f}'\left( x \right)$
$\displaystyle \int{{kf\left( x \right)\ dx}}=\ \ k\int{{f\left( x \right)\ dx}}$
$ \displaystyle \frac{d}{{dx}}\left[ {f\left( x \right)\pm g\left( x \right)} \right]=\frac{d}{{dx}}f\left( x \right)\pm \frac{d}{{dx}}g\left( x \right)$
$ \displaystyle \int{{\left[ {f\left( x \right)\pm g\left( x \right)} \right]\ dx}}= \int{{f\left( x \right)\ dx}}\pm \int{{g\left( x \right)\ dx}}$
အထက်ပါ ထောင့်မှန်စတုဂံများ၏ ဧရိယာများပေါင်းလဒ်ကို Riemann sum ဟုခေါ်သည်။
အောက်ပါ geogebra applet ဖြင့်လေ့လာကြည့်ပါ။
applet တွင် တွေ့ရှိချက်အရ left sum, right sum တို့ထက် midpoint sum သည် အဖြေမှန်နှင့် ပိုမို နီးစပ်သည်ကို တွေ့ရှိရပေမည်။ သို့ရာတွင် စိပ်ပိုင်းသည့် အရေအတွက် အနန္တ (∞) သို့ ချဉ်းကပ်သွား သည့်အခါ left sum နှင့် right sum တို့သည့်ကွားခြားမှု မရှိတော့ပေ။
If $ \displaystyle f$ is continuous and nonnegative on the closed interval $ \displaystyle [a, b]$, then the area of the region bounded by the graph of $ \displaystyle f$, the $ \displaystyle x$-axis, and the vertical lines $ \displaystyle x = a$ and $ \displaystyle x = b$ is $ \displaystyle A$, then
