Tentukan integral tak tentu berikut: e.

Pertanyaan

Tentukan integral tak tentu berikut:

e. $integral open parentheses 2 x squared minus x plus 3 close parentheses d x$

Pembahasan Soal:

Rumus dasar integral yaitu:

$integral k x to the power of n space d x equals fraction numerator k over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus straight C$ dengan sayarat $straight n not equal to negative 1$
$integral k space d x equals k x plus straight C$ , suatu konstanta

Diperoleh penyelesaiannya yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses 2 x squared minus x plus 3 close parentheses d x end cell equals cell integral 2 x squared space d x minus integral x space d x plus integral 3 space d x end cell row blank equals cell fraction numerator 2 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent minus fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent plus 3 x plus straight C end cell row blank equals cell 2 over 3 x cubed minus 1 half x squared plus 3 x plus straight C end cell end table$

Dengan demikian, integral tak tentu dari $integral open parentheses 2 x squared minus x plus 3 close parentheses d x$ adalah $2 over 3 x cubed minus 1 half x squared plus 3 x plus straight C$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Ayu

Mahasiswa/Alumni Universitas Muhammadiyah Prof. DR. Hamka

Terakhir diupdate 11 Juli 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

Hitunglah pengintegralan di bawah ini! 5)

Pembahasan Soal:

Integral fungsi $begin mathsize 14px style f left parenthesis x right parenthesis equals a x to the power of n end style$ dapat ditentukan sebagai berikut.

$begin mathsize 14px style integral space a x to the power of n space d x equals fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C end style$

sehingga integral dari fungsi yang diberikan di atas dapat ditentukan sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral space 8 x cubed minus 3 x squared plus x plus 5 space straight d x end cell equals cell 8 over 4 x to the power of 4 minus 3 over 3 x cubed plus 1 half x squared plus 5 x plus C end cell row blank equals cell 2 x to the power of 4 minus x cubed plus 1 half x squared plus 5 x plus C end cell end table end style$

Dengan demikian, hasil integral fungsi yang diberikan adalah $Error converting from MathML to accessible text.$ .

Roboguru

Hasil dari adalah ....

Pembahasan Soal:

Untuk menentukan hasil integral dari $integral open parentheses 2 minus x close parentheses squared open parentheses 3 x plus 1 close parentheses space d x$ , perhatikan perhitungan berikut.

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses 2 minus x close parentheses squared open parentheses 3 x plus 1 close parentheses space d x end cell row blank equals cell integral open parentheses 3 x cubed minus 11 x squared plus 8 x plus 4 close parentheses blank d x end cell row blank equals cell integral 3 x cubed blank d x minus integral 11 x squared blank d x plus integral 8 x blank d x plus integral 4 blank d x end cell row blank equals cell 3 open parentheses fraction numerator 1 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent plus C subscript 1 close parentheses minus 11 open parentheses fraction numerator 1 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus C subscript 2 close parentheses plus end cell row blank blank cell 8 open parentheses fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent plus C subscript 3 close parentheses plus 4 open parentheses x plus C subscript 4 close parentheses end cell row blank equals cell 3 open parentheses 1 fourth x to the power of 4 plus C subscript 1 close parentheses minus 11 open parentheses 1 third x cubed plus C subscript 2 close parentheses plus end cell row blank blank cell 8 open parentheses 1 half x squared plus C subscript 3 close parentheses plus 4 open parentheses x plus C subscript 4 close parentheses end cell row blank equals cell 3 over 4 x to the power of 4 plus 3 C subscript 1 minus 11 over 3 x cubed minus 11 C subscript 2 plus 4 x squared plus 8 C subscript 3 plus 4 x plus 4 C subscript 4 end cell row blank equals cell 3 over 4 x to the power of 4 minus 11 over 3 x cubed plus 4 x squared plus 4 x plus 3 C subscript 1 minus 11 C subscript 2 plus 8 C subscript 3 plus 4 C subscript 4 end cell end table end style$

Misalkan $3 C subscript 1 minus 11 C subscript 2 plus 8 C subscript 3 plus 4 C subscript 4 equals C$ , maka didapat hasil perhitungan sebagai berikut.

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 3 over 4 x to the power of 4 minus 11 over 3 x cubed plus 4 x squared plus 4 x plus 3 C subscript 1 minus 11 C subscript 2 plus 8 C subscript 3 plus 4 C subscript 4 end cell row blank equals cell 3 over 4 x to the power of 4 minus 11 over 3 x cubed plus 4 x squared plus 4 x plus C end cell end table end style$

Jadi, jawaban yang tepat adalah A.

Roboguru

Hitunglah pengintegralan di bawah ini! 4)

Pembahasan Soal:

Integral fungsi $begin mathsize 14px style f left parenthesis x right parenthesis equals a x to the power of n end style$ dapat ditentukan sebagai berikut.

$begin mathsize 14px style integral space a x to the power of n space d x equals fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C end style$

sehingga integral dari fungsi yang diberikan di atas dapat ditentukan sebagai berikut.

$begin mathsize 14px style integral space 2 x squared minus 3 x minus 2 space straight d x equals 2 over 3 x cubed minus 3 over 2 x cubed minus 2 x plus C end style$

Roboguru

Hasil dari adalah ....

Pembahasan Soal:

Perhatikan perhitungan berikut!

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral 1 half open parentheses x plus 1 close parentheses open parentheses 1 minus x close parentheses squared blank straight d x end cell row blank equals cell integral 1 half open parentheses x cubed minus x squared minus x plus 1 close parentheses blank straight d x end cell row blank equals cell 1 half integral open parentheses x cubed minus x squared minus x plus 1 close parentheses blank straight d x end cell row blank equals cell 1 half open parentheses integral x cubed blank straight d x minus integral x squared blank straight d x minus integral x blank straight d x plus integral 1 blank straight d x close parentheses end cell row blank equals cell 1 half open parentheses open parentheses fraction numerator 1 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent plus C subscript 1 close parentheses minus open parentheses fraction numerator 1 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus C subscript 2 close parentheses minus open parentheses fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent plus C subscript 3 close parentheses plus open parentheses x plus C subscript 4 close parentheses close parentheses end cell row blank equals cell 1 half open parentheses open parentheses 1 fourth x to the power of 4 plus C subscript 1 close parentheses minus open parentheses 1 third x cubed plus C subscript 2 close parentheses minus open parentheses 1 half x squared plus C subscript 3 close parentheses plus open parentheses x plus C subscript 4 close parentheses close parentheses end cell row blank equals cell 1 half open parentheses 1 fourth x to the power of 4 plus C subscript 1 minus 1 third x cubed minus C subscript 2 minus 1 half x squared minus C subscript 3 plus x plus C subscript 4 close parentheses end cell row blank equals cell 1 over 8 x to the power of 4 plus C subscript 1 over 2 minus 1 over 6 x cubed minus C subscript 2 over 2 minus 1 fourth x squared minus C subscript 3 over 2 plus x over 2 plus C subscript 4 over 2 end cell row blank equals cell 1 over 8 x to the power of 4 minus 1 over 6 x cubed minus 1 fourth x squared plus x over 2 plus C subscript 1 over 2 minus C subscript 2 over 2 minus C subscript 3 over 2 plus C subscript 4 over 2 end cell end table end style$

Jika $begin mathsize 12px style C subscript 1 over 2 minus C subscript 2 over 2 minus C subscript 3 over 2 plus C subscript 4 over 2 equals C comma end style$ maka diperoleh hasil sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 over 8 x to the power of 4 minus 1 over 6 x cubed minus 1 fourth x squared plus x over 2 plus C subscript 1 over 2 minus C subscript 2 over 2 minus C subscript 3 over 2 plus C subscript 4 over 2 end cell row blank equals cell 1 over 8 x to the power of 4 minus 1 over 6 x cubed minus 1 fourth x squared plus x over 2 plus C end cell end table$

Jadi, jawaban yang tepat adalah B.

Roboguru

Tentukan integral tak tentu berikut: g.

Pembahasan Soal:

Rumus dasar integral yaitu:

$integral k x to the power of n space d x equals fraction numerator k over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus straight C$ dengan sayarat $straight n not equal to negative 1$
$integral k space d x equals k x plus straight C$ , suatu konstanta

Diperoleh penyelesaiannya yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses 2 x minus 1 over x close parentheses squared space d x end cell equals cell integral open parentheses 2 x minus 1 over x close parentheses open parentheses 2 x minus 1 over x close parentheses d x end cell row blank equals cell integral open parentheses 4 x squared minus 2 minus 2 plus 1 over x squared close parentheses d x end cell row blank equals cell integral open parentheses 4 x squared minus 4 plus 1 over x squared close parentheses d x end cell row blank equals cell integral 4 x squared space d x minus integral 4 space d x plus integral 1 over x squared space d x end cell row blank equals cell equals integral 4 x squared space d x minus integral 4 space d x plus integral x to the power of negative 2 end exponent space d x end cell row blank equals cell fraction numerator 4 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent minus 4 x plus fraction numerator 1 over denominator negative 2 plus 1 end fraction x to the power of negative 2 plus 1 end exponent plus straight C end cell row blank equals cell 4 over 3 x cubed minus 4 x plus fraction numerator 1 over denominator negative 1 end fraction x to the power of negative 1 end exponent plus straight C end cell row blank equals cell 4 over 3 x cubed minus 4 x minus x to the power of negative 1 end exponent plus straight C end cell row blank equals cell 4 over 3 x cubed minus 4 x minus 1 over x plus straight C end cell end table$

Dengan demikian, integral tak tentu dari $integral open parentheses 2 x minus 1 over x close parentheses squared space d x$ adalah $Error converting from MathML to accessible text.$ .

Roboguru