Exercise Zone : Luas Daerah Di Bawah Kurva

Berikut ini adalah kumpulan soal mengenai Luas Daerah Di Bawah Kurva. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Luas daerah yang diarsir adalah .... satuan luas.

$$\begin{array}{rl}L& =-{\displaystyle \underset{1}{\overset{2}{\int }}(-x^2-1)dx}\\ & =-{[-{\displaystyle \frac{1}{3}}{x}^3-x]}_1^2\\ & =-[(-{\displaystyle \frac{1}{3}}(2{)}^3-2)-(-{\displaystyle \frac{1}{3}}(1{)}^3-1)]\\ & =-[(-{\displaystyle \frac{8}{3}}-2)-(-{\displaystyle \frac{1}{3}}-1)]\\ & =-[-{\displaystyle \frac{14}{3}}-(-{\displaystyle \frac{4}{3}})]\\ & =-[-{\displaystyle \frac{10}{3}}]\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{10}{3}}}}}}\end{array}$$

Jadi, luas daerah yang diarsir adalah \dfrac{10}3 satuan luas. JAWAB: B

No.

Luas daerah yang diarsir adalah

Titik potong kurva terhadap sumbu x
$$\begin{array}{rl}-4x^3+8x^2& =0\\ -4x^2(x-2)& =0\end{array}$$
x=0 dan x=2

$$\begin{array}{rl}L& ={\displaystyle \underset{0}{\overset{2}{\int }}(-4x^3+8x^2)}\\ & ={[-x^4+{\displaystyle \frac{8}{3}}{x}^3]}_0^2\\ & =[-2^4+{\displaystyle \frac{8}{3}}(2{)}^3]-[-0^4+{\displaystyle \frac{8}{3}}(0{)}^3]\\ & =[-16+{\displaystyle \frac{8}{3}}(8)]-[0+{\displaystyle \frac{8}{3}}(0)]\\ & =[-16+{\displaystyle \frac{64}{3}}]-\left[0\right]\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{16}{3}}}}}}\end{array}$$

Jadi, luas daerah yang diarsir adalah \dfrac{16}3 satuan luas. JAWAB: B

No.

Luas daerah yang dibatasi oleh kurva y=x^2+1, x=2, sumbu y, dan sumbu x adalah .... satuan luas.

$$\begin{array}{rl}L& ={\displaystyle \underset{0}{\overset{2}{\int }}(x^2+1)dx}\\ & ={[{\displaystyle \frac{1}{3}}{x}^3+x]}_0^2\\ & =[{\displaystyle \frac{1}{3}}(2{)}^3+2]-[{\displaystyle \frac{1}{3}}(0{)}^3+0]\\ & =[{\displaystyle \frac{1}{3}}(8)+2]-[0+0]\\ & =[{\displaystyle \frac{8}{3}}+2]-0\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{14}{3}}}}}}\end{array}$$

Jadi, luasnya adalah \dfrac{14}3 satuan luas. JAWAB: E

No.

Luas daerah yang diarsir pada gambar di samping adalah

kita bagi menjadi dua daerah seperti pada gambar berikut:

Luas daerah I
$$\begin{array}{rl}{L}_1& ={\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x}dx}\\ & ={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{\frac{1}{2}}dx}\\ & ={\left[{\displaystyle \frac{2}{3}}{x}^{\frac{3}{2}}\right]}_0^1\\ & ={\displaystyle \frac{2}{3}}{\left[x\sqrt{x}\right]}_0^1\\ & ={\displaystyle \frac{2}{3}}[1\sqrt{1}-0\sqrt{0}]\\ & ={\displaystyle \frac{2}{3}}\end{array}$$

Luas daerah 2
$$\begin{array}{rl}{L}_2& ={\displaystyle \frac{1}{2}}\cdot 1\cdot 1\\ & ={\displaystyle \frac{1}{2}}\end{array}$$

Luas keseluruhan,
$$\begin{array}{rl}L& =L_1+L_2\\ & ={\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{7}{6}}}}}}\end{array}$$

Jadi, luasnya adalah \dfrac76 satuan luas. JAWAB: C

Luas daerah yang diarsir pada gambar adalah

$$\begin{array}{rl}L& ={\displaystyle \underset{0}{\overset{4}{\int }}(2x-(x^2-2x))dx}\\ & ={\displaystyle \underset{0}{\overset{4}{\int }}(2x-x^2+2x)dx}\\ & ={\displaystyle \underset{0}{\overset{4}{\int }}(-x^2+4x)dx}\\ & ={[-{\displaystyle \frac{1}{3}}{x}^3+2x^2]}_0^4\\ & =[-{\displaystyle \frac{1}{3}}(4{)}^3+2(4{)}^2]-[-{\displaystyle \frac{1}{3}}(0{)}^3+(0{)}^2]\\ & =[-{\displaystyle \frac{64}{3}}+32]-0\\ & =-21{\displaystyle \frac{1}{3}}+32\\ & =11-{\displaystyle \frac{1}{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 10{\displaystyle \frac{2}{3}}}}}}\end{array}$$

Jadi, luasnya adalah 10\dfrac23 satuan luas. JAWAB: B

No.

Luas daerah antara kurva y=-x^3-x^2+2x dengan sumbu x adalah

Titik potong kurva terhadap sumbu x
$$\begin{array}{rl}-x^3-x^2+2x& =0\\ -x(x^2+x-2)& =0\\ -x(x+2)(x-1)& =0\end{array}$$
x=0, x=-2, x=1

Ada 2 daerah, yaitu dari x=-2 sampai x=0, dan dari x=0 sampai x=1.

Luas daerah 1

karena daerahnya berada di bawah sumbu x, maka kita gunakan tanda negatif sebelum tanda integral.
$$\begin{array}{rl}{L}_1& =-{\displaystyle \underset{-2}{\overset{0}{\int }}(-x^3-x^2+2x)dx}\\ & =-{[-{\displaystyle \frac{1}{4}}{x}^4-{\displaystyle \frac{1}{3}}{x}^3+x^2]}_{-2}^0\\ & =-[(-{\displaystyle \frac{1}{4}}(0{)}^4-{\displaystyle \frac{1}{3}}(0{)}^3+0^2)-(-{\displaystyle \frac{1}{4}}(-2{)}^4-{\displaystyle \frac{1}{3}}(-2{)}^3+(-2{)}^2)]\\ & =-[(-{\displaystyle \frac{1}{4}}(0)-{\displaystyle \frac{1}{3}}(0)+0)-(-{\displaystyle \frac{1}{4}}(16)-{\displaystyle \frac{1}{3}}(-8)+4)]\\ & =-[(0-0+0)-(-4+{\displaystyle \frac{8}{3}}+4)]\\ & =-[0-{\displaystyle \frac{8}{3}}]\\ & ={\displaystyle \frac{8}{3}}\end{array}$$

Luas daerah 2

$$\begin{array}{rl}{L}_1& ={\displaystyle \underset{0}{\overset{1}{\int }}(-x^3-x^2+2x)dx}\\ & ={[-{\displaystyle \frac{1}{4}}{x}^4-{\displaystyle \frac{1}{3}}{x}^3+x^2]}_0^1\\ & =[-{\displaystyle \frac{1}{4}}(1{)}^4-{\displaystyle \frac{1}{3}}(1{)}^3+1^2]-[-{\displaystyle \frac{1}{4}}(0{)}^4-{\displaystyle \frac{1}{3}}(0{)}^3+0^2]\\ & =[-{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{3}}+1]-\left[0\right]\\ & ={\displaystyle \frac{5}{12}}\end{array}$$

Luas keseluruhan

$$\begin{array}{rl}L& =L_1+L_2\\ & ={\displaystyle \frac{8}{3}}+{\displaystyle \frac{5}{12}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{37}{12}}}}}}\end{array}$$

Jadi, luasnya adalah \dfrac{37}{12} satuan luas. JAWAB: C

No.

Luas daerah yang dibatasi oleh kurva y=x^2-4x+3, sumbu x dan sumbu y adalah .... satuan luas.

Titik potong kurva dengan sumbu x
$$\begin{array}{rl}{x}^2-4x+3& =0\\ (x-1)(x-3)& =0\end{array}$$
x=1 dan x=3

$$\begin{array}{rl}L& ={\displaystyle \underset{0}{\overset{1}{\int }}(x^2-4x+3)dx}\\ & ={[{\displaystyle \frac{1}{3}}{x}^3-2x^2+3x]}_0^1\\ & =[{\displaystyle \frac{1}{3}}(1{)}^3-2(1{)}^2+3(1)]-[{\displaystyle \frac{1}{3}}(0{)}^3-2(0{)}^2+3(0)]\\ & ={\displaystyle \frac{1}{3}}-2+3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{3}}}}}}\end{array}$$

Jadi, luasnya adalah \dfrac43 satuan luas. JAWAB: C

No.

Luas daerah yang diarsir pada gambar di samping ini adalah .... satuan luas.

CARA 1

Fungsi kuadrat tersebut mempunyai titik puncak di (3,0).
$$\begin{array}{rl}y& =a{(x-x_p)}^2+y_p\\ & =a{(x-3)}^2+0\\ & =a{(x-3)}^2\end{array}$$

Melalui (0,2),
$$\begin{array}{rl}2& =a{(0-3)}^2\\ 2& =a{(-3)}^2\\ 2& =a\left(9\right)\\ a& ={\displaystyle \frac{2}{9}}\end{array}$$
Substitusikan ke persamaan awal,
y=\dfrac29(x-3)^2

Luas daerahnya,
$$\begin{array}{rl}L& ={\displaystyle \underset{0}{\overset{3}{\int }}({\displaystyle \frac{2}{9}}(x-3{)}^2)dx}\\ & ={[{\displaystyle \frac{2}{9}}\cdot {\displaystyle \frac{1}{3}}(x-3{)}^3]}_0^3\\ & ={\displaystyle \frac{2}{27}}{[(x-3{)}^3]}_0^3\\ & ={\displaystyle \frac{2}{27}}[(3-3{)}^3-(0-3{)}^3]\\ & ={\displaystyle \frac{2}{27}}[0^3-(-3{)}^3]\\ & ={\displaystyle \frac{2}{27}}[0-(-27)]\\ & ={\displaystyle \frac{2}{27}}[0+27]\\ & ={\displaystyle \frac{2}{27}}\left[27\right]\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$$

CARA 2

$$\begin{array}{rl}L& ={\displaystyle \frac{2\cdot 3}{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$$

Jadi, luasnya adalah 2 satuan luas. JAWAB: A

No.

Luas daerah yang diarsir pada gambar di bawah ini adalah ....

SKB CPNS MATEMATIKA 2020

CARA BIASA

Titik Potong Kedua Kurva

$\begin{array}{rl}3x-x^2& =-x+3\\ 3x-x^2+x-3& =0\\ -x^2+4x-3& =0\\ x^2-4x+3& =0\\ (x-1)(x-3)& =0\end{array}$
x=1 dan x=3

Luas Daerah

$\begin{array}{rl}L& ={\displaystyle \underset{1}{\overset{3}{\int }}(3x-x^2)-(-x+3)dx}\\ & ={\displaystyle \underset{1}{\overset{3}{\int }}3x-x^2+x-3dx}\\ & ={\displaystyle \underset{1}{\overset{3}{\int }}-x^2+4x-3dx}\\ & ={[-{\displaystyle \frac{1}{3}}{x}^3+2x^2-3x]}_1^3\\ & =[-{\displaystyle \frac{1}{3}}(3{)}^3+2(3{)}^2-3(3)]-[-{\displaystyle \frac{1}{3}}(1{)}^3+2(1{)}^2-3(1)]\\ & =[-9+18-9]-[-{\displaystyle \frac{1}{3}}+2-3]\\ & =[0]-[-1{\displaystyle \frac{1}{3}}]\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1{\displaystyle \frac{1}{3}}}}}}\end{array}$

CARA CEPAT

$\begin{array}{rl}3x-x^2-(-x+3)& =3x-x^2+x-3\\ & =-x^2+4x-3\end{array}$
a=-1, b=4, c=-3

$\begin{array}{rl}D& =b^2-4ac\\ & =4^2-4(-1)(-3)\\ & =16-12\\ & =4\end{array}$

$\begin{array}{rl}L& ={\displaystyle \frac{D\sqrt{D}}{6a^2}}\\ & ={\displaystyle \frac{4\sqrt{4}}{6(-1{)}^2}}\\ & ={\displaystyle \frac{8}{6}}\\ & ={\displaystyle \frac{4}{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1{\displaystyle \frac{1}{3}}}}}}\end{array}$

Jadi, luasnya adalah 1\dfrac13 satuan luas.

No.

Sketch the curve and find the total area between the curve and the given interval on the x-axis
y=\sqrt{x+1}-1; [-1,1]

brainly.co.id

$\begin{array}{rl}{L}_1& =-{\displaystyle \underset{-1}{\overset{0}{\int }}(\sqrt{x+1}-1)dx}\\ & =-{\displaystyle \underset{-1}{\overset{0}{\int }}((x+1{)}^{\frac{1}{2}}-1)d(x+1)}\\ & =-{[{\displaystyle \frac{2}{3}}(x+1{)}^{\frac{3}{2}}-x]}_{-1}^0\\ & =-[({\displaystyle \frac{2}{3}}(0+1{)}^{\frac{3}{2}}-0)-({\displaystyle \frac{2}{3}}(-1+1{)}^{\frac{3}{2}}-(-1))]\\ & =-[({\displaystyle \frac{2}{3}}(1{)}^{\frac{3}{2}})-({\displaystyle \frac{2}{3}}(0{)}^{\frac{3}{2}}+1)]\\ & =-[{\displaystyle \frac{2}{3}}-1]\\ & =-[-{\displaystyle \frac{1}{3}}]\\ & ={\displaystyle \frac{1}{3}}\end{array}$

$\begin{array}{rl}{L}_2& ={\displaystyle \underset{0}{\overset{1}{\int }}(\sqrt{x+1}-1)dx}\\ & ={\displaystyle \underset{0}{\overset{1}{\int }}((x+1{)}^{\frac{1}{2}}-1)d(x+1)}\\ & ={[{\displaystyle \frac{2}{3}}(x+1{)}^{\frac{3}{2}}-x]}_0^1\\ & =({\displaystyle \frac{2}{3}}(1+1{)}^{\frac{3}{2}}-1)-({\displaystyle \frac{2}{3}}(0+1{)}^{\frac{3}{2}}-0)\\ & =({\displaystyle \frac{2}{3}}(2{)}^{\frac{3}{2}}-1)-({\displaystyle \frac{2}{3}}(1{)}^{\frac{3}{2}})\\ & =({\displaystyle \frac{2}{3}}\left(2\sqrt{2}\right)-1)-\left({\displaystyle \frac{2}{3}}\right)\\ & ={\displaystyle \frac{4}{3}}\sqrt{2}-1-{\displaystyle \frac{2}{3}}\\ & ={\displaystyle \frac{4}{3}}\sqrt{2}-{\displaystyle \frac{5}{3}}\end{array}$

{L=L_1+L_2=\dfrac13+\dfrac43\sqrt2-\dfrac53=\boxed{\boxed{\dfrac43\sqrt2-\dfrac43}}}

So, the total area between the curve and the given interval on the x-axis is \dfrac43\sqrt2-\dfrac43

LIHAT JUGA:
INTEGRAL TENTU INTEGRAL TAK TENTU