Konsep dan Rumus Integral Serta Pembahasan Soal

KOMSEP DAN RUMUS INTEGRAL

Pada postingan sebelumnya sudah dibahas tentang integral tak tentu dan postingan ini akan membahas integral tentu yang dimulai dari teorema dasar kalkulus yang diajarkan si tingkat SMA/MA/SMK/MAK, dimana teorema ini penting dalam menyediakan alat ampuh untuk perhitungan integral tentu. Dan yang paling membuatnya penting terletak pada hubungan yang terbentuk antara turunan dan integral. Untuk lebih jelasnya simak pembahasan beberapa teorema dasar berikut.

Teorema A : Teorema Dasar Kalkulus

Andaikan f kontinu sehingga terintegralkan pada selang [a, b] dan andaikan F sembarang antiturunan dari f pada selang tersebut, maka

$\int_{a}^{b}f(x)dx=F(b)-F(a)$

Teorema B : Teorema Kelinearan Integral Tentu

Andaikan bahwa f dan g terintegralkan pada selang [a, b] dan k konstanta, maka kf dan f + g adalah terintegralkan serta

$\int_{a}^{b}kf(x)dx=k\int_{a}^{b}f(x)dx$

$\int_{a}^{b}\left [ f(x)+g(x) \right ]dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$

$\int_{a}^{b}\left [ f(x)-g(x) \right ]dx=\int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx$

Contoh Soal dan Pembahasan

Hitunglah integral berikut.

1. $\int_{0}^{2}x^{3}dx$

Pembahasan

Gunakan Teorema A

$\int_{0}^{2}x^{3}dx=\left [ \frac{x^{4}}{4} \right ]_{0}^{2}$

$=\left ( \frac{2^{4}}{4} \right )-\left ( \frac{0^{4}}{4} \right )$

$=\frac{16}{4}=4$

2. $\int_{-1}^{2}x^{4}dx$

Pembahasan

$\int_{-1}^{2}x^{4}dx=\left [ \frac{x^{5}}{5} \right ]_{-1}^{2}$

$=\left ( \frac{2^{5}}{5} \right )-\left ( \frac{(-1)^{5}}{5} \right )$

$=\left ( \frac{32}{5} \right )-\left ( \frac{-1}{5} \right )$

$=\frac{32}{5}+\frac{1}{5}$

$=\frac{33}{5}$

3. $\int_{1}^{2}\left ( 4x^{3}+7 \right )dx$

Pembahasan

$\int_{1}^{2}\left ( 4x^{3}+7 \right )dx=\int_{1}^{2}4x^{3}+\int_{1}^{2}7dx$

$=\left [ x^{4}\right ]_{1}^{2}+\left [ 7x \right ]_{1}^{2}$

$=\left ( 2^{4}-1^{4} \right )+\left ( 7.2-7.1 \right )$

$=(16-1)+(14-7)$

$=15-7=8$

4. $\int_{-1}^{2}\left ( 3x^{2}-2x+3 \right )dx$

Pembahasan

$\int_{-1}^{2}\left ( 3x^{2}-2x+3 \right )dx=\left [ x^{3}-x^{2}+3x \right ]_{-1}^{2}$

$=\left ( 2^{3}-2^{2}+3.2 \right )-\left ( (-1)^{3}-(-1)^{2}+3(-1) \right )$

$=\left ( 8-4+6 \right )-\left ( -1-1-3 \right )$

$=10+5=15$

5. $\int_{1}^{4}\frac{1}{w^{2}}dw$

Penjelasan

$\int_{1}^{4}\frac{1}{w^{2}}dw=\int_{1}^{4}w^{-2}dw$

$=\left [ -w^{-1} \right ]_{1}^{4}$

$=((-4^{-1})-(-1^{-1}))$

$=-\frac{1}{4}-\left ( -1 \right )$

$=-\frac{1}{4}+1=\frac{3}{4}$

6. $\int_{1}^{3}\frac{2}{t^{2}}dt$

Pembahasan

$\int_{1}^{3}\frac{2}{t^{2}}dt=2\int_{1}^{3}t^{-2}dt$

$=2\left [ -t^{-1} \right ]_{1}^{3}$

$=\left [ \frac{-2}{t} \right ]_{1}^{3}$

$=\left ( \frac{-2}{3} \right )-\left ( -2 \right )$

$=\frac{-2}{3}+2$

$=\frac{4}{3}$

7. $\int_{0}^{4}\sqrt{t}dt$

Pembahasan

$\int_{0}^{4}\sqrt{t}dt=\int_{0}^{4}t^{\frac{1}{2}}dt$

$=\left [ \frac{2}{3}t^{\frac{3}{2}} \right ]_{0}^{4}$

$=\frac{2}{3}4^{\frac{3}{2}}-0$

$=\frac{2}{3}\left ( 2^{2} \right )^{\frac{3}{2}}$

$=\frac{2}{3}2^{3}$

$=\frac{2}{3}.8$

$=\frac{16}{3}=5\frac{1}{3}$

8. $\int_{1}^{8}\sqrt[3]{w}dw$

Pembahasan

$\int_{1}^{8}\sqrt[3]{w}dw=\int_{1}^{8}w^{\frac{1}{3}}dw$

$=\left [ \frac{3}{4}w^{\frac{4}{3}}\right]_{1}^{8}$

$=\left ( \frac{3}{4}8^{\frac{4}{3}} \right )-\left ( \frac{3}{4}1^{\frac{4}{3}} \right )$

$=\frac{3}{4}2^{4}-\frac{3}{4}$

$=\frac{3}{4}16-\frac{3}{4}$

$=\frac{3}{4}(16-1)$

$=\frac{3}{4}(15)$

$=\frac{45}{4}=11\frac{1}{4}$

9. $\int_{-4}^{-1}\left ( y^{2}+\frac{1}{y^{3}} \right )dy$

Pembahasan

$\int_{-4}^{-1}\left ( y^{2}+\frac{1}{y^{3}} \right )dy=\int_{-4}^{-1}y^{2}dy+\int_{-4}^{-1}\frac{1}{y^{3}}dy$

$=\int_{-4}^{-1}y^{2}dy+\int_{-4}^{-1}y^{-3}dy$

$=\left [ \frac{1}{3}y^{3} \right ]_{-4}^{-1}+\left [ -\frac{1}{2}y^{-2} \right ]_{-4}^{-1}$

$=\left [ \left ( \frac{1}{3}\left ( -1 \right )^{3} \right )-\left ( \frac{1}{3}\left ( -4 \right )^{3} \right ) \right ]+\left [ \left ( -\frac{1}{2}\left ( -1 \right )^{-2} \right )-\left ( -\frac{1}{2}\left ( -4 \right )^{-2} \right ) \right ]$

$=\left [ \left ( -\frac{1}{3} \right )-\left ( \frac{1}{3}\left ( -64 \right ) \right ) \right ]+\left [ \left ( -\frac{1}{2} \right )-\left ( -\frac{1}{2}\frac{1}{16} \right ) \right ]$

$=\left [ \left ( -\frac{1}{3} \right )+\left ( \frac{64}{3} \right ) \right ]+\left [ \left ( -\frac{1}{2} \right )+\left ( \frac{1}{32} \right ) \right ]$

$=\frac{63}{3}+\left [ \left ( -\frac{16}{32} \right )+\left ( \frac{1}{32} \right ) \right ]$

$=\frac{63}{3}-\frac{15}{32}$

$=21-\frac{15}{32}$

$=\frac{672-15}{32}$

$=\frac{657}{32}=2\frac{17}{32}$

10. $\int_{1}^{4}\frac{s^{4}-8}{s^{2}}ds$

Pembahasan

$\int_{1}^{4}\frac{s^{4}-8}{s^{2}}ds=\int_{1}^{4}\left ( s^{2}-\frac{8}{s^{2}} \right )ds$

$=\int_{1}^{4}\left ( s^{2}-8s^{-2} \right )ds$

$=\left [ \frac{1}{3}s^{3}+8s^{-1} \right ]_{1}^{4}$

$=\left ( \frac{1}{3}4^{3}+8(4)^{-1} \right )-\left ( \frac{1}{3}1^{3}+8(1)^{-1} \right )$

$=\left ( \frac{64}{3}+2 \right )-\left ( \frac{1}{3}+8 \right )$

$=\frac{70}{3}-\frac{25}{3}$

$=\frac{45}{3}=15$

11. $\int_{0}^{1}\left ( x^{2}+1 \right )^{10}\left ( 2x \right )dx$

Pembahasan

Ingat bahwa $\int\left ( g(x) \right )^{n}g'(x)dx=\frac{\left [ g(x) \right ]^{n+1}}{n+1}+c$ . Untuk lebih jelasnya dapat dilihat disini

Misal $g(x)=x^{2}+1\rightarrow g'(x)=2x$ , maka

$\int_{0}^{1}\left ( x^{2}+1 \right )^{10}\left ( 2x \right )dx=\left [ \frac{\left ( x^{2}+1 \right )^{11}}{11} \right ]_{0}^{1}$

$=\left [ \frac{\left ( 1^{2}+1 \right )^{11}}{11} \right ]-\left [ \frac{\left ( 0^{2}+1 \right )^{11}}{11} \right ]$

$=\frac{2^{11}}{11}-\frac{1}{11}$

$=\frac{2^{11}-1}{11}$

$=\frac{2048-1}{11}$

$=\frac{2047}{11}=186\frac{1}{11}$

12. $\int_{-1}^{0}\sqrt{x^{3}+1}\left ( 3x^{2} \right )dx$

Pembahasan

$\int_{-1}^{0}\sqrt{x^{3}+1}\left ( 3x^{2} \right )dx=\int_{-1}^{0}\left ( x^{3}+1 \right )^{\frac{1}{2}}\left ( 3x^{2} \right )dx$

Misal $g(x)=x^{3}+1 \rightarrow g'(x)=3x^{2}$ , maka

$=\left [ \frac{\left ( x^{3}+1 \right )^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{-1}^{0}$

$=\left [ \frac{2\left ( x^{3}+1 \right )^{\frac{3}{2}}}{3} \right ]_{-1}^{0}$

$=\left [ \frac{2\left ( 0^{3}+1 \right )^{\frac{3}{2}}}{3} \right ]-\left [ \frac{2\left ( (-1)^{3}+1 \right )^{\frac{3}{2}}}{3} \right ]$

$=\left [ \frac{2\left ( 1 \right )^{\frac{3}{2}}}{3} \right ]-\left [ \frac{2\left ( 0 \right )^{\frac{3}{2}}}{3} \right ]$

$=\left [ \frac{2\left ( 1 \right )}{3} \right ]-\left [ \frac{2\left ( 0 \right )}{3} \right ]$

$=\frac{2}{3}$

Demikian pembahasan konsep integral tentu disertai rumus dan pembahasan soal. Semoga dapat dipahami dan bermanfaat.

Konsep dan Rumus Integral Serta Pembahasan Soal

KOMSEP DAN RUMUS INTEGRAL

Contoh Soal dan Pembahasan

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Konsep dan Rumus Integral Serta Pembahasan Soal

KOMSEP DAN RUMUS INTEGRAL

Contoh Soal dan Pembahasan

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