Inetgral Tentu Fungsi Aljabar

A. Integral Tentu

Sebelumnya sudah dibahasa mengenai integral tak tentu, di mana integral ini memiliki ciri selalu ada nilai konstantanya. Hal ini menunjukkan bahwa ada nilai yang belum terukur dengan jelas. Jika nantinya pada integral sudah ditentukan batas bawah dan batas atasnya, sehingga tidak akan diperlukan lagi nilai konstantanya. Adapun rumus formulasi dari integral tentu adalah sebagai berikut:

$\displaystyle \int_{a}^{b}f(x)\: dx=\left [ F(x) \right ]_{a}^{b}=F(x)|_{a}^{b}=F(b)-F(a)$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{aligned}1.\quad &\int_{1}^{3}\left ( 2x-1 \right )\: dx\\ &=\left [ x^{2}-x \right ]_{1}^{3}\\ &=(9-3)-(1-1)=6\\ 2.\quad &\int_{0}^{5}\left | x-1 \right |+\left | x-2 \right |\: dx\\ &=\frac{1}{2}(x-1)\left | x-1 \right ||_{0}^{5}\: +\frac{1}{2}(x-2)\left | x-2 \right ||_{0}^{5}\\ &=\left ( \frac{1}{2}.4.4+\frac{1}{2}.3.3 \right )-\left ( \frac{1}{2}.-1.1+\frac{1}{2}.-2.2 \right )\\ &=8+4\frac{1}{2}+\frac{1}{2}+2=15 \end{aligned}$.

Anda dapat mengecek jawaban di atas dengan melmperhatikan gambar kemudian menyelesaikannya dengan prinsip segitiga, persegi, dan atau persegipanjang.

B. Sifat-Sifat Integral Tentu

$\begin{aligned}1.\quad&\int_{a}^{b}f(x)\: dx=-\int_{b}^{a}f(x)\: dx\\ 2.\quad&\displaystyle \int_{a}^{b}k.f(x)\: dx=k\int_{a}^{b}f(x)\: dx\\ 3.\quad&\displaystyle \int_{a}^{b}\left ( f(x)\pm g(x) \right )\: dx\\ &=\displaystyle \int_{a}^{b}f(x)\: dx\: \pm\int_{a}^{b}g(x) \: dx\\ 4.\quad &\displaystyle \int_{a}^{b}f(x)\: dx\\ &=\displaystyle \int_{a}^{c}f(x)\: dx+\int_{c}^{b}f(x)\: dx\: ,\: \: dengan\: \: a<c<b\\ 5.\quad &\displaystyle \int_{a}^{a}f(x)\: dx=0 \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Hitunglah hasil dari integral}\: \: \displaystyle \int_{1}^{3}\left ( 2y+1 \right )\: dy\\ \\\\&\textbf{Jawab}:\\ &\begin{aligned}\int_{1}^{3}\left ( 2y+1 \right )\: dy&=y^{2}+y\: |_{1}^{3}\\ &=\left ( 3^{2}+3 \right )-\left ( 1^{2}+1 \right )\\ &=12-2\\ &=10 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Hitunglah hasil dari integral}\: \:  \displaystyle \int_{-1}^{1}x^{3}-1\: dx\\\\&\textbf{Jawab}:\\ &\begin{aligned}\int_{-1}^{1}&\left ( x^{3}-1 \right )\: dx\\ &=-(\displaystyle \frac{1}{4}x^{4}-x|_{-1}^{1})\\ &=-\left ( \displaystyle \frac{1}{4}.\left ( 1 \right )^{4}-1 \right )+\left ( \displaystyle \frac{1}{4}.\left ( -1 \right )^{4}-\left ( -1 \right ) \right )\\ &=-\left ( \displaystyle \frac{1}{4}-1 \right )+\left ( \displaystyle \frac{1}{4}+1 \right )\\ &=2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Hitunglah hasil dari integral}\: \:  \displaystyle \int_{-2}^{1}\left ( x+3 \right )^{2}\: dx\\\\&\textbf{Jawab}:\\ &\begin{aligned}\int_{-2}^{1}&\left ( x+3 \right )^{2}\: dx\\ &=\displaystyle \frac{1}{3}x^{2}+3x^{2}+9x|_{-2}^{1}\\ &=\left ( \displaystyle \frac{1}{3}.\left ( 1 \right )^{3}+3.1^{2}+9.1 \right )-\left ( \displaystyle \frac{1}{3}.\left ( -2 \right )^{3}+3.(-2)^{2}+9.(-2) \right )\\ &=\left ( \displaystyle \frac{1}{3}+12\right )-\left ( -\displaystyle \frac{8}{3}+12-18 \right )\\ &=\displaystyle \frac{9}{3}+18=3+18=21 \end{aligned} \end{array}$ .

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

$\begin{array}{ll}\\ &\textrm{Hitunglah nilai tiap integral berikut ini}\\ &\begin{array}{lllll}\\ 1.&\displaystyle \int_{0}^{3}2x\: dx&10.&\displaystyle \int_{1}^{2}\sqrt{x^{5}}\: dx\\ 2.&\displaystyle \int_{2}^{3}2x^{2}\: dx&11.&\displaystyle \int_{1}^{2}\displaystyle \frac{1}{x^{2}}\: dx\\ 3.&\displaystyle \int_{3}^{7}\displaystyle \frac{1}{2}x^{3}\: dx&12.&\displaystyle \int_{4}^{9}3\sqrt{x}\: dx\\ 4.&\displaystyle \int_{0}^{3}\left ( x^{2}+2x-1 \right )\: dx&\displaystyle \\ 5.&\displaystyle \int_{0}^{3}\left ( x+3 \right )^{2}\: dx&\\ 6.&\displaystyle \displaystyle \int_{-1}^{2}\left ( x+2 \right )\left ( x-1 \right )\: dx&\\ 7.&\displaystyle \int_{1}^{3}\left ( t+t^{2}-3t^{3} \right )\: dt&\\ 8.&\displaystyle \int_{1}^{4}\left ( \displaystyle \frac{x^{4}-x^{3}+\sqrt{x}-1}{x^{2}} \right )\: dx&\\ 9.&\displaystyle \int_{4}^{6}\left ( \sqrt{t}+t^{2}-\sqrt[3]{t^{2}} \right )\: dt \end{array}  \end{array}$.