Roboguru

Nilai dari ∫−11​(31​x2+3x−2)dx=....

Pertanyaan

Nilai dari integral subscript negative 1 end subscript superscript 1 open parentheses 1 third x squared plus 3 x minus 2 close parentheses d x equals.... 

  1. negative 34 over 9 

  2. negative 32 over 9 

  3. negative 2 over 9 

  4. 2 over 9 

  5. 34 over 9 

Pembahasan Soal:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral subscript negative 1 end subscript superscript 1 open parentheses 1 third x squared plus 3 x minus 2 close parentheses d x end cell row blank equals cell open square brackets 1 third times fraction numerator 1 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus fraction numerator 3 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent minus fraction numerator 2 over denominator 0 plus 1 end fraction x to the power of 0 plus 1 end exponent close square brackets subscript negative 1 end subscript superscript 1 end cell row blank equals cell open square brackets 1 over 9 x cubed plus 3 over 2 x squared minus 2 x close square brackets subscript negative 1 end subscript superscript 1 end cell row blank equals cell open parentheses 1 over 9 open parentheses 1 close parentheses cubed plus 3 over 2 open parentheses 1 close parentheses squared minus 2 open parentheses 1 close parentheses close parentheses minus end cell row blank blank cell open parentheses 1 over 9 open parentheses negative 1 close parentheses cubed plus 3 over 2 open parentheses negative 1 close parentheses squared minus 2 open parentheses negative 1 close parentheses close parentheses end cell row blank equals cell open parentheses 1 over 9 plus up diagonal strike 3 over 2 end strike minus 2 close parentheses minus open parentheses negative 1 over 9 plus up diagonal strike 3 over 2 end strike plus 2 close parentheses end cell row blank equals cell 2 over 9 minus 4 end cell row blank equals cell fraction numerator 2 minus 36 over denominator 9 end fraction end cell row blank equals cell fraction numerator negative 34 over denominator 9 end fraction end cell row blank equals cell negative 34 over 9 end cell end table end style     

Jadi, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Nur

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Jika  dengan  merupakan fungsi genap dan  merupakan fungsi ganjil, maka hasil dari adalah ...

Pembahasan Soal:

Perhatikan perhitungan berikut!

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript negative 3 end subscript superscript 3 f open parentheses x close parentheses open parentheses g open parentheses x close parentheses minus 1 close parentheses space d x end cell equals cell integral subscript negative 3 end subscript superscript 3 open square brackets f open parentheses x close parentheses g open parentheses x close parentheses minus f open parentheses x close parentheses close square brackets space d x end cell row 6 equals cell integral subscript negative 3 end subscript superscript 3 f open parentheses x close parentheses g open parentheses x close parentheses space d x minus integral subscript negative 3 end subscript superscript 3 f open parentheses x close parentheses space d x end cell end table end style

Diketajui undefined merupakan fungsi genap dan undefined merupakan fungsi ganjil. Akibatnya, didapat hubungan berikut ini.

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses negative x close parentheses end cell equals cell f open parentheses x close parentheses end cell row cell g open parentheses negative x close parentheses end cell equals cell negative g open parentheses x close parentheses end cell end table

Misal begin mathsize 14px style h open parentheses x close parentheses equals f open parentheses x close parentheses g open parentheses x close parentheses end style, maka didapat hasil perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell h open parentheses negative x close parentheses end cell equals cell f open parentheses negative x close parentheses g open parentheses negative x close parentheses end cell row blank equals cell negative f open parentheses x close parentheses g open parentheses x close parentheses end cell row blank equals cell negative h open parentheses x close parentheses end cell end table end style

Karena h open parentheses negative x close parentheses equals negative h open parentheses x close parentheses,maka  undefined merupakan fungsi ganjil.

Oleh karena itu, didapat perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row 6 equals cell integral subscript negative 3 end subscript superscript 3 f open parentheses x close parentheses g open parentheses x close parentheses space d x minus integral subscript negative 3 end subscript superscript 3 f open parentheses x close parentheses space d x end cell row 6 equals cell 0 minus 2 integral subscript 0 superscript 3 f open parentheses x close parentheses space d x end cell row 6 equals cell negative 2 integral subscript 0 superscript 3 f open parentheses x close parentheses space d x end cell row cell negative 3 end cell equals cell integral subscript 0 superscript 3 f open parentheses x close parentheses space d x end cell end table

Dengan demikian, hasil dariintegral subscript 0 superscript 3 f open parentheses x close parentheses space d x adalah negative 3.

Oleh karena itu, jawaban yang tepat adalah B.

0

Roboguru

∫−22​(3x2+4)dx=....

Pembahasan Soal:

integral subscript negative 2 end subscript superscript 2 left parenthesis 3 x squared space plus space 4 right parenthesis d x space equals space open square brackets 1 third. space 3 x cubed space plus space 4. x close square brackets subscript negative 2 end subscript superscript 2  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open square brackets x cubed space plus space 4 x close square brackets subscript negative 2 end subscript superscript 2  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open square brackets left parenthesis 2 right parenthesis cubed space plus space 4 left parenthesis 2 right parenthesis close square brackets space minus space open square brackets left parenthesis negative 2 right parenthesis cubed space plus space 4 left parenthesis negative 2 right parenthesis close square brackets  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 16 space plus space 16  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 32

0

Roboguru

Fungsi f(x) memenuhi f(x)=f(−x). Jika nilai ∫−33​f(x)dx=6 dan ∫23​f(x)dx=1, maka nilai ∫02​f(x)dx=…

Pembahasan Soal:

Diketahui fungsi f left parenthesis x right parenthesis memenuhi f left parenthesis x right parenthesis equals f left parenthesis negative x right parenthesis merupakan fungsi genap yang memenuhi sifat integral subscript negative a end subscript superscript a f left parenthesis x right parenthesis space d x equals 2 integral subscript 0 superscript a f left parenthesis x right parenthesis space d x.

Secara umum, bentuk integral juga dapat diubah batasnya seperti berikut, integral subscript a superscript c f left parenthesis x right parenthesis space d x equals integral subscript a superscript b f left parenthesis x right parenthesis space d x plus integral subscript b superscript c f left parenthesis x right parenthesis space d x untuk rentang a less or equal than b less or equal than c.

Diberikan nilai integral integral subscript negative 3 end subscript superscript 3 f left parenthesis x right parenthesis space d x equals 6, maka bentuknya dapat diubah menjadi 

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript negative 3 end subscript superscript 3 f left parenthesis x right parenthesis space d x end cell equals 6 row cell 2 integral subscript 0 superscript 3 f left parenthesis x right parenthesis space d x end cell equals 6 row cell integral subscript 0 superscript 3 f left parenthesis x right parenthesis space d x end cell equals cell 6 over 2 end cell row cell integral subscript 0 superscript 3 f left parenthesis x right parenthesis space d x end cell equals 3 end table 

Diketahui pula nilai  integral subscript 2 superscript 3 f left parenthesis x right parenthesis space d x equals 1, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 3 f left parenthesis x right parenthesis space d x end cell equals cell integral subscript 0 superscript 2 f left parenthesis x right parenthesis space d x plus integral subscript 2 superscript 3 f left parenthesis x right parenthesis space d x end cell row cell integral subscript 0 superscript 2 f left parenthesis x right parenthesis space d x end cell equals cell integral subscript 0 superscript 3 f left parenthesis x right parenthesis space d x minus integral subscript 2 superscript 3 f left parenthesis x right parenthesis space d x end cell row blank equals cell 3 minus 1 end cell row blank equals 2 end table 

Jadi, nilai dari integral subscript 0 superscript 2 f left parenthesis x right parenthesis space d x equals 2.

0

Roboguru

Jika ∫−44​ f(x) (sin x + 1) dx = 8, dengan f(x) fungsi genap dan ∫−24​ f(x) dx = 4, maka ∫−20​ f(x) dx = ....

Pembahasan Soal:

begin mathsize 14px style Ada space beberapa space hal space yang space perlu space diingat space colon  Untuk space fungsi space ganjil space colon space straight f left parenthesis short dash straight a right parenthesis equals short dash straight f left parenthesis straight a right parenthesis space berlaku  integral from negative straight p to straight p of space straight f open parentheses straight x close parentheses dx equals 0    Untuk space fungsi space genap space colon space straight f left parenthesis short dash straight a right parenthesis equals straight f left parenthesis straight a right parenthesis space berlaku  integral from negative straight p to straight p of space straight f open parentheses straight x close parentheses dx equals 2 integral subscript 0 superscript straight p space straight f open parentheses straight x close parentheses dx    left parenthesis fungsi space genap right parenthesis left parenthesis space fungsi space ganjil right parenthesis equals fungsi space ganjil  left parenthesis fungsi space genap right parenthesis left parenthesis fungsi space genap right parenthesis equals fungsi space genap    Kurva space straight y equals sin invisible function application space straight x space adalah space fungsi space ganjil.  integral subscript negative 4 end subscript superscript 4 space straight f open parentheses straight x close parentheses open parentheses sin space invisible function application straight x plus 1 close parentheses dx equals 8  integral subscript negative 4 end subscript superscript 4 stack stack space straight f open parentheses straight x close parentheses space sin space invisible function application straight x space with underbrace below with fungsi straight space ganjil below dx plus integral subscript negative 4 end subscript superscript 4 stack stack straight f open parentheses straight x close parentheses with underbrace below with fungsi straight space genap below dx equals 8  0 plus 2 integral subscript 0 superscript 4 space straight f open parentheses straight x close parentheses space dx equals 8  integral subscript 0 superscript 4 space straight f open parentheses straight x close parentheses space dx equals 4    integral subscript negative 2 end subscript superscript 4 space straight f open parentheses straight x close parentheses space dx equals 4  integral subscript negative 2 end subscript superscript 0 space straight f open parentheses straight x close parentheses straight space dx plus integral subscript 0 superscript 4 space straight f open parentheses straight x close parentheses straight space dx equals 4  integral subscript negative 2 end subscript superscript 0 space straight f open parentheses straight x close parentheses straight space dx plus 4 equals 4  integral subscript negative 2 end subscript superscript 0 space straight f open parentheses straight x close parentheses straight space dx equals 0 end style

0

Roboguru

Diketahui f(x) adalah fungsi genap. Jika ∫−55​f(x)dx=14dan∫15​f(x)dx=10, maka ∫−11​f(x)dx=....

Pembahasan Soal:

Diketahui begin mathsize 14px style f open parentheses x close parentheses end style adalah fungsi genap, berlaku begin mathsize 14px style integral subscript negative a end subscript superscript a f left parenthesis x right parenthesis d x equals 2 integral subscript 0 superscript a f left parenthesis x right parenthesis d x end style

Ingat: begin mathsize 14px style integral subscript a superscript c f left parenthesis x right parenthesis space d x equals integral subscript a superscript b f left parenthesis x right parenthesis space d x space plus integral subscript b superscript c f left parenthesis x right parenthesis space d x space dengan space a less than b less than c end style

Sehingga

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript negative 5 end subscript superscript 5 f left parenthesis x right parenthesis d x end cell equals 14 row cell 2 integral subscript 0 superscript 5 f left parenthesis x right parenthesis d x end cell equals 14 row cell integral subscript 0 superscript 5 f left parenthesis x right parenthesis d x end cell equals 7 row cell integral subscript 0 superscript 1 f left parenthesis x right parenthesis d x plus integral subscript 1 superscript 5 f left parenthesis x right parenthesis d x end cell equals 7 row cell integral subscript 0 superscript 1 f left parenthesis x right parenthesis d x plus 10 end cell equals 7 row cell integral subscript 0 superscript 1 f left parenthesis x right parenthesis d x end cell equals cell negative 3 end cell end table end style

Diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript negative 1 end subscript superscript 1 f left parenthesis x right parenthesis d x end cell equals cell 2 integral subscript 0 superscript 1 f left parenthesis x right parenthesis d x end cell row blank equals cell 2 times open parentheses negative 3 close parentheses end cell row blank equals cell negative 6 end cell end table end style

Jadi, jawaban yang tepat adalah B.
 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved