Tag: Limit

Menurut teorema integral tak tentu, definisi integral tak wajar, dan konsep limit tak hingga, berlaku
$\begin{aligned} \displaystyle \int_1^{\infty} x^{-2}~\text{d}x & = \lim_{b \to \infty} \int_1^b x^{-2}~\text{d}x \\ & = \lim_{b \to \infty} \left[-\dfrac{1}{x}\right]_1^b \\ & = \lim_{b \to \infty} \left(1-\dfrac{1}{b}\right) = 1. \end{aligned}$
Jadi, integral tersebut konvergen ke $\boxed{1}$

Menurut teorema integral tak tentu, definisi integral tak wajar, dan konsep limit tak hingga, berlaku
$$\begin{aligned} \displaystyle \int_1^{\infty} \dfrac{1}{(1+x)^2}~\text{d}x & = \lim_{b \to \infty} \int_1^b \dfrac{1}{(1+x)^2}~\text{d}x \\ & = \lim_{b \to \infty} \left[-\dfrac{1}{1+x}\right]_1^b \\ & = \lim_{b \to \infty} \left(\dfrac{1}{2}-\dfrac{1}{1+b}\right) = \dfrac{1}{2}. \end{aligned}$$Jadi, integral tersebut konvergen ke $\boxed{\dfrac{1}{2}}$

Menurut teorema integral tak tentu, definisi integral tak wajar, dan konsep limit tak hingga, berlaku
$$\begin{aligned} \displaystyle \int_{-\infty}^{1} e^{2x}~\text{d}x & = \lim_{a \to-\infty} \int_a^1 e^{2x}~\text{d}x \\ & = \lim_{a \to-\infty} \left[\dfrac{1}{2}e^{2x}\right]_a^1 \\ & = \lim_{a \to-\infty} \left(\dfrac{1}{2}e^2- \dfrac{1}{2}e^{2a}\right) \\ & = \dfrac{1}{2} \lim_{a \to-\infty} e^2- \dfrac{1}{2} \lim_{a \to-\infty} e^{2a} \\ & = \dfrac{1}{2}e^2- \dfrac{1}{2}(0)= \dfrac{1}{2}e^2. \end{aligned}$$Jadi, integral tersebut konvergen ke $\boxed{\dfrac{1}{2}e^2}$

Ingat bahwa $\int a^{nx}~\text{d}x = \dfrac{a^x}{n \ln a} + C$, dengan syarat $a > 0$ dan $a \neq 1$. 
Menurut teorema integral tak tentu, definisi integral tak wajar, dan konsep limit tak hingga, berlaku
$\begin{aligned} \displaystyle \int_{-\infty}^{0} 3^{8x}~\text{d}x & = \lim_{b \to-\infty} \int_b^0 3^{8x}~\text{d}x \\ & = \dfrac{1}{8 \ln 3} \lim_{b \to-\infty} [3^{8x}]_b^0 \\ & = \dfrac{1}{8 \ln 3} \lim_{b \to-\infty} (3^0- 3^{8b}) \\ & = \dfrac{1}{8 \ln 3}(1- 0) \\ & = \dfrac{1}{8 \ln 3}. \end{aligned}$
Jadi, integral tersebut konvergen ke $\boxed{\dfrac{1}{8 \ln 3}}$

Ingat bahwa 
$$\boxed{\displaystyle \int \dfrac{1}{a^2+u^2}~\text{d}u = \dfrac{1}{a} \arctan \left(\dfrac{u}{a}\right)+ C}$$Menurut teorema integral tak tentu, definisi integral tak wajar, dan konsep limit tak hingga, berlaku
$$\begin{aligned} \displaystyle \int_{-\infty}^{\infty} \dfrac{1}{1+x^2}~\text{d}x & = \int_{-\infty}^{0} \dfrac{1}{1+x^2}~\text{d}x + \int_{0}^{\infty} \dfrac{1}{1+x^2}~\text{d}x \\ & = \lim_{a \to-\infty} \int_{a}^{0} \dfrac{1}{1+x^2}~\text{d}x + \lim_{b \to \infty} \int_{0}^{b} \dfrac{1}{1+x^2}~\text{d}x \\ & = \lim_{a \to-\infty} [\arctan x]_a^0 + \lim_{b \to \infty} [\arctan x]_0^b \\ & = \left(0 + \dfrac{\pi} {2}\right) + \left(\dfrac{\pi}{2} + 0\right) = \pi. \end{aligned}$$Jadi, integral tersebut konvergen ke $\boxed{\pi}$

Menurut teorema integral tak tentu, definisi integral tak wajar, dan konsep limit tak hingga, berlaku
$$\begin{aligned} \displaystyle \int_{-\infty}^{\infty} xe^{-x^2}~\text{d}x & = \int_{-\infty}^{0} xe^{-x^2}~\text{d}x + \int_{0}^{\infty} xe^{-x^2}~\text{d}x \\ & = \lim_{a \to-\infty} \int_{a}^{0} xe^{-x^2}~\text{d}x + \lim_{b \to \infty} \int_{0}^{b} xe^{-x^2}~\text{d}x \\ & = \lim_{a \to-\infty} \left[-\dfrac{1}{2}e^{-x^2}\right]_a^0 + \lim_{b \to \infty} \left[-\dfrac{1}{2}e^{-x^2}\right]_0^b \bigstar \\ & = \lim_{a \to-\infty} \left(-\dfrac{1}{2} + \dfrac{1}{2}e^{-a^2}\right) + \lim_{b \to \infty} \left(\dfrac{1}{2}- \dfrac{1}{2}e^{-b^2}\right) \\ & =-\dfrac{1}{2} + 0 +\dfrac{1}{2}- 0 = 0. \end{aligned}$$Jadi, integral tersebut konvergen ke $\boxed{0}$ 
$\bigstar$ Cara mengintegralkan bentuk di atas adalah sebagai berikut.
Misalkan $u =-x^2$, berarti $\text{d}u =-2x~\text{d}x$ sehingga
$\begin{aligned} \int xe^{-x^2}~\text{d}x & =-\dfrac{1}{2} \int e^u~\text{d}u \\ & =-\dfrac{1}{2}e^u + C \\ &  =-\dfrac{1}{2}e^{-x^2} + C \end{aligned}$

Menurut teorema integral tak tentu, definisi integral tak berhingga di suatu titik ujung, dan konsep limit, berlaku
$\begin{aligned} \displaystyle \int_0^1 x^{-2}~\text{d}x & = \lim_{u \to 0^+} \int_u^1 \dfrac{1}{x^2}~\text{d}x \\ & = \lim_{u \to 0^+} \left[-\dfrac{1}{x}\right]_u^1 \\ & = \lim_{u \to 0^+} \left(-1 + \dfrac{1}{u}\right) = \infty. \end{aligned}$
Karena limitnya tak hingga, maka integral tersebut divergen.

Menurut teorema integral tak tentu, definisi integral tak berhingga di suatu titik ujung, dan konsep limit, berlaku
$\begin{aligned} \displaystyle \int_0^1 \dfrac{\ln x}{x}~\text{d}x & = \lim_{u \to 0^+} \ln x~\text{d}(\ln x) \\ & = \dfrac{1}{2} \lim_{u \to 0^+} [\ln^2 x]_u^1 \\ & = \dfrac{1}{2} \lim_{u \to 0^+} (0- \ln^2 u) \\ & = \infty. \end{aligned}$
Karena limitnya tak hingga, maka integral tersebut divergen.