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In Matematika / Sekolah Menengah Atas | 2025-07-09

1. Hitunglah luas daerah tertutup yang dibatasi kurva y = x2,sumbu Y,dan garis y = 5
2. Hitunglah luas daerah tertutup yang dibatasi kurva y = 4x - x2, sumbu x, dan garis x =8
3. Hitunglah luas daerah di kuadran I yang dibatasi kurva y2 = x, garis x + y = 4, dan sumbu x

Asked by mastyozdv93361

Answer (3)

Arranging the following numbers in increasing order (2.3, 45.1, 18.735, 0.9862, 7
) would be:


9862



3


7


735,



1



Take note of the decimal points in which these numbers are associated with however, since these given numbers total varies in interval between each other, this is not directly observed unlike 1.2 1.5 and 1.0

Answered by taskmasters | 2024-06-10

The numbers arranged in increasing order are 0.9862, 2.3, 7, 18.735, and 45.1. We compare the values of each number to determine their positions. This list shows the numbers from smallest to largest clearly.
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Answered by taskmasters | 2024-12-26

Jawab:Penjelasan dengan langkah-langkah:1.[tex]y=x^2[/tex][tex]x=0\to y=0\\y=5\\x^2=5\\x=\pm\sqrt{5}[/tex][tex]L=5\sqrt{5} -\int\limits^{\sqrt{5} }_0 {x^2} \, dx \\L=5\sqrt{5} -\int\limits^{\sqrt{5} }_0 {x^2} \, dx \\L=5\sqrt{5} -\frac{1}{3}x^{3} \,\,|^{\sqrt{5}}_0 \\L=5\sqrt{5} -(\frac{1}{3}(\sqrt{5} )^3 -0)\\L=5\sqrt{5}-\frac{5}{3} \sqrt{5}\\L=\frac{15}{3} \sqrt{5}-\frac{5}{3} \sqrt{5}\\L=\frac{10}{3} \sqrt{5}[/tex]2.[tex]y=4-x^2[/tex][tex]y=0\\4x-x^2=0\\x(4-x)=0\\x=0\text{ atau }x=4[/tex][tex]L=\int\limits^4_0 {4x-x^2} \, dx -\int\limits^8_4 {4x-x^2} \, dx\\L=[2x^2-\frac{1}{3} x^3 ]_0^4-[2x^2-\frac{1}{3} x^3 ]_4^8\\L=((2(4)^2-\frac{1}{3}(4)^3)-(2(0)^2-\frac{1}{3}(0)^3)-((2(8)^2-\frac{1}{3}(8)^3)-(2(4)^2-\frac{1}{3}(4)^3)\\L=(32-\frac{64}{3})-((128-\frac{512}{3})-(32-\frac{64}{3})\\L=2(32-\frac{64}{3})-(128-\frac{512}{3})\\L=64-\frac{128}{3} -128+\frac{512}{3} \\L=-64+\frac{384}{3} \\L=-64+128\\L=64[/tex]3.[tex]y^2=x\\y=\pm\sqrt{x} \\[/tex][tex]x+y=4\\y=4-x[/tex]nilai x pada titik potong  kedua kurva di kuadran I:[tex]\sqrt{x} =4-x\\x=4^2-8x+x^2\\x^2-9x+16=0\\x=\frac{9\pm\sqrt{(-9)^2-4(1)(16)} }{2} \\x=\frac{9\pm\sqrt{81-64} }{2} \\x=\frac{9\pm\sqrt{17} }{2}[/tex]karena di kuadran I maka:[tex]x=\frac{9+\sqrt{17} }{2}[/tex]Penyelesaian untuk x tidak sederhana sehingga integrasinya akan panjang sekali. Silakan  khusus nomor 3 dibuat pertanyaan terpisah agar saya atau rekan lain bisa membantu.

Answered by mastreeada | 2025-07-10