Benang Lurus
Benang Lurus
Integral Aljabar
Integral Tak Tentu
$$\bbox[black, 5px, border: 2px solid red] {\int ax^n \text{ dx} = \dfrac{a}{n + 1} x^{n + 1} + C}$$
Contoh 01
\(\int 2x^3 \text{ dx} = \dfrac{2}{3 + 1} x^{3 + 1} + C\)
\(\int 2x^3 \text{ dx} = \dfrac{1}{2} x^{4} + C\)
Contoh 02
\(\int \sqrt[3]{x^2} \text{ dx} = \int x^{\frac{2}{3}} \text{ dx}\)
\(\int \sqrt[3]{x^2} \text{ dx} = \dfrac{1}{\frac{2}{3} + 1} x^{\frac{2}{3} + 1} + C\)
\(\int \sqrt[3]{x^2} \text{ dx} = \dfrac{1}{\frac{5}{3}} x^{\frac{2}{3}}\cdot x^{1} + C\)
\(\int \sqrt[3]{x^2} \text{ dx} = \frac{3}{5} x \sqrt[3]{x^2}+ C\)
Contoh 03
\(\int \dfrac{x^2 + 1}{x} \text{ dx} = \int \left(\dfrac{x^2}{x} + \dfrac{1}{x}\right) \text{ dx}\)
\(\int \dfrac{x^2 + 1}{x} \text{ dx} = \int \left(x + \dfrac{1}{x}\right) \text{ dx}\)
\(\int \dfrac{x^2 + 1}{x} \text{ dx} = \dfrac{1}{1 + 1} x^{1 + 1} + \ln |x| + C\)
Note: \(\color{cyan}\int \frac{1}{x} \text{ dx} = \ln {|x|} + C\)
\(\int \dfrac{x^2 + 1}{x} \text{ dx} = \dfrac{1}{2} x^{2} + \ln |x| + C\)
Contoh 04
\(\int 5 \text{ dx} = 5x + C\)
\(\int 7 \text{ dx} = 7x + C\)
\(\int p \text{ dx} = px + C\)
Integral Substitusi
Contoh 01
\(\int (2x + 1)\sqrt{2x^2 + 2x} \text{ dx}\)
Misal \(u = 2x^2 + 2x\) maka \(\dfrac{\text{du}}{\text{dx}} = 4x + 2\)
dapat ditulis \(\text{dx}= \dfrac{\text{du}}{2(2x + 1)}\)
\(\int (2x + 1)\sqrt{2x^2 + 2x} \text{ dx} = \int \cancel{(2x + 1)} \sqrt{u} \cdot \dfrac{\text{du}}{2\cancel{(2x + 1)}}\)
\(\int (2x + 1)\sqrt{2x^2 + 2x} \text{ dx} = \dfrac{1}{2}\cdot \int u^{\frac{1}{2}}\text{ du} \)
\(\int (2x + 1)\sqrt{2x^2 + 2x} \text{ dx} = \dfrac{1}{2}\cdot \dfrac{1}{\frac{1}{2} + 1} \cdot u^{\frac{1}{2} + 1} + C \)
\(\int (2x + 1)\sqrt{2x^2 + 2x} \text{ dx} = \dfrac{1}{2}\cdot \dfrac{2}{3} \cdot u^{\frac{1}{2}}\cdot u^{1} + C \)
\(\int (2x + 1)\sqrt{2x^2 + 2x} \text{ dx} = \dfrac{1}{3} \cdot u\sqrt{u} + C \)
\(\int (2x + 1)\sqrt{2x^2 + 2x} \text{ dx} = \dfrac{1}{3} \cdot (2x^2 + 2x)\sqrt{2x^2 + 2x} + C \)
Contoh 02
\(\int \dfrac{6x}{(3x^2 + 1)^5}\text{ dx}\)
Misal \(u = 3x^2 + 1\) maka \(\dfrac{\text{du}}{\text{dx}} = 6x\)
dapat ditulis \(\text{dx}= \dfrac{\text{du}}{6x}\)
\(\int \dfrac{6x}{(3x^2 + 1)^5}\text{ dx} = \int \dfrac{\cancel{6x}}{u^5} \cdot \dfrac{\text{du}}{\cancel{6x}}\)
\(\int \dfrac{6x}{(3x^2 + 1)^5}\text{ dx} = \int \dfrac{1}{u^5} \text{ du}\)
\(\int \dfrac{6x}{(3x^2 + 1)^5}\text{ dx} = \int u^{-5} \text{ du}\)
\(\int \dfrac{6x}{(3x^2 + 1)^5}\text{ dx} = \dfrac{1}{-5 + 1} \cdot u^{-5 + 1} + C\)
\(\int \dfrac{6x}{(3x^2 + 1)^5}\text{ dx} = -\dfrac{1}{4} \cdot (3x^2 + 1)^{-4} + C\)
\(\int \dfrac{6x}{(3x^2 + 1)^5}\text{ dx} = -\dfrac{1}{4(3x^2 + 1)^4} + C\)
Integral Tentu
$$\bbox[black, 5px, border: 2px solid red] {\int_{a}^{b} f(x) \text{ dx} = F(a)\:-\:F(b)}$$
Contoh 01
\(\int_{-1}^{2} (2x + 1) \text{ dx} = (x^2 + x)\:\:|_{-1}^{2}\)
\(\int_{-1}^{2} (2x + 1) \text{ dx} = (2^2 + 2)\:-\:((-1)^2 \:-\:1)\)
\(\int_{-1}^{2} (2x + 1) \text{ dx} = 6\:-\:0 = 0\)
Contoh 02
\(\int_{2}^{7} \dfrac{1}{\sqrt{x + 2}}\text{ dx} = \int_{2}^{7} (x + 2)^{-\frac{1}{2}}\text{ dx}\)
\(\int_{2}^{7} \dfrac{1}{\sqrt{x + 2}}\text{ dx} = \dfrac{1}{-\frac{1}{2} + 1} (x + 2)^{-\frac{1}{2} + 1} \:\:|_{2}^{7}\)
\(\int_{2}^{7} \dfrac{1}{\sqrt{x + 2}}\text{ dx} = 2(x + 2)^{\frac{1}{2}} \:\:|_{2}^{7}\)
\(\int_{2}^{7} \dfrac{1}{\sqrt{x + 2}}\text{ dx} = 2\sqrt{x + 2}\:\:|_{2}^{7}\)
\(\int_{2}^{7} \dfrac{1}{\sqrt{x + 2}}\text{ dx} = 2\sqrt{7 + 2}\:-\:2\sqrt{2 + 2}\)
\(\int_{2}^{7} \dfrac{1}{\sqrt{x + 2}}\text{ dx} = 2(3)\:-\:2(2)\)
\(\int_{2}^{7} \dfrac{1}{\sqrt{x + 2}}\text{ dx} = 2\)