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Selesaikan masing-masing integral parsial berikut ini dengan formula: . a. ∫xx−16​dx

Pertanyaan

Selesaikan masing-masing integral parsial berikut ini dengan formula:

begin mathsize 14px style integral straight u times dv equals straight u times straight v space minus space integral straight v space du end style.

a. xx16dx 

Pembahasan Soal:

Misalkan:

u=xdv=x16dv=(x16)21dxdu=1du=dxv=(x16)21dx=(21+1)1(x16)21+1=231(x16)23=32(x16)23  

Sehingga diperoleh:

xx16dx=====uvvdux(32(x16)23)(x16)23dx32x(x16)23(23+1)1(x16)23+1+C32x(x16)2352(x16)25+C32x(x16)(x16)52(x16)2(x16)+C 

Dengan demikian, hasil dari:

Error converting from MathML to accessible text.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Tessalonika

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 30 Agustus 2021

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Pertanyaan yang serupa

Tentukan hasil dari integral berikut: g. Tentukan nilai perhitungan integral parsial

Pembahasan Soal:

integral x open parentheses x minus 3 close parentheses to the power of 6 d x

Perhatikan rumus integral parsial berikut ini.

integral U space d V equals U times V minus integral V space d U

Misalkan

table attributes columnalign right center left columnspacing 0px end attributes row U equals cell x space rightwards arrow d U equals d x end cell row cell d V end cell equals cell open parentheses x minus 3 close parentheses to the power of 6 d x rightwards arrow V equals integral open parentheses x minus 3 close parentheses to the power of 6 d x equals 1 over 7 open parentheses x minus 3 close parentheses to the power of 7 end cell end table

Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell integral U space d V end cell equals cell U times V minus integral V space d U end cell row cell integral x open parentheses x minus 3 close parentheses to the power of 6 d x end cell equals cell x times 1 over 7 open parentheses x minus 3 close parentheses to the power of 7 minus integral 1 over 7 open parentheses x minus 3 close parentheses to the power of 7 d x end cell row blank equals cell 1 over 7 x open parentheses x minus 3 close parentheses to the power of 7 minus 1 over 7 times 1 over 8 open parentheses x minus 3 close parentheses to the power of 8 plus C end cell row blank equals cell 1 over 7 x open parentheses x minus 3 close parentheses to the power of 7 minus 1 over 56 open parentheses x minus 3 close parentheses to the power of 8 plus C end cell end table

Dengan demikian hasil dari

Error converting from MathML to accessible text..

Roboguru

Pembahasan Soal:

Gunakan konsep integral parsial.

integral u space straight d v equals u v minus integral v space straight d u integral k open parentheses a x plus b close parentheses to the power of n d x equals k over a times fraction numerator 1 over denominator n plus 1 end fraction open parentheses a x plus b close parentheses to the power of n plus 1 end exponent plus c

Pilih u equals x dan straight d v equals open parentheses x plus 1 close parentheses to the power of 5 space straight d x.

u equals x rightwards arrow fraction numerator straight d u over denominator straight d x end fraction equals 1 rightwards arrow straight d u equals straight d x

table attributes columnalign right center left columnspacing 2px end attributes row cell straight d v end cell equals cell open parentheses x plus 1 close parentheses to the power of 5 space straight d x end cell row v equals cell integral d v end cell row blank equals cell integral open parentheses x plus 1 close parentheses to the power of 5 space straight d x end cell row blank equals cell 1 over 1 times fraction numerator 1 over denominator 5 plus 1 end fraction open parentheses x plus 1 close parentheses to the power of 5 plus 1 end exponent end cell row v equals cell 1 over 6 open parentheses x plus 1 close parentheses to the power of 6 end cell end table

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell integral u space d v end cell equals cell u v minus integral v d u end cell row cell integral x open parentheses x plus 1 close parentheses to the power of 5 space straight d x end cell equals cell x times 1 over 6 open parentheses x plus 1 close parentheses to the power of 6 minus integral 1 over 6 open parentheses x plus 1 close parentheses to the power of 6 space d x end cell row blank equals cell x over 6 open parentheses x plus 1 close parentheses to the power of 6 minus 1 over 6 integral open parentheses x plus 1 close parentheses to the power of 6 space d x end cell row blank equals cell x over 6 open parentheses x plus 1 close parentheses to the power of 6 minus 1 over 6 times fraction numerator 1 over denominator 6 plus 1 end fraction open parentheses x plus 1 close parentheses to the power of 6 plus 1 end exponent plus c end cell row blank equals cell x over 6 open parentheses x plus 1 close parentheses to the power of 6 minus 1 over 42 open parentheses x plus 1 close parentheses to the power of 7 plus c end cell row blank equals cell x over 6 open parentheses x plus 1 close parentheses to the power of 6 minus fraction numerator open parentheses x plus 1 close parentheses over denominator 42 end fraction open parentheses x plus 1 close parentheses to the power of 6 plus c end cell row blank equals cell fraction numerator 7 x minus open parentheses x plus 1 close parentheses over denominator 42 end fraction open parentheses x plus 1 close parentheses to the power of 6 plus c end cell row blank equals cell fraction numerator open parentheses 6 x minus 1 close parentheses over denominator 42 end fraction open parentheses x plus 1 close parentheses to the power of 6 plus c end cell row cell integral x open parentheses x plus 1 close parentheses to the power of 5 space straight d x end cell equals cell 1 over 42 open parentheses x plus 1 close parentheses to the power of 6 open parentheses 6 x minus 1 close parentheses plus c end cell end table

Jadi, jawaban yang tepat adalah A.

Roboguru

Tentukan integral berikut dengan integral parsial!

Pembahasan Soal:

Bentuk umum integral parsial yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral f open parentheses x close parentheses times g open parentheses x close parentheses d x end cell equals cell integral u space d v end cell row blank equals cell u v minus integral v space d u end cell end table

Diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell x cubed end cell row cell fraction numerator d u over denominator d x end fraction end cell equals cell 3 x squared end cell row cell d u end cell equals cell 3 x squared d x end cell end table

dan

table attributes columnalign right center left columnspacing 0px end attributes row cell d v end cell equals cell open parentheses x plus 1 close parentheses cubed end cell row v equals cell integral open parentheses x plus 1 close parentheses cubed x end cell row blank equals cell fraction numerator 1 over denominator 3 plus 1 end fraction open parentheses x plus 1 close parentheses to the power of 3 plus 1 end exponent end cell row blank equals cell 1 fourth open parentheses x plus 1 close parentheses to the power of 4 end cell end table

Substitusi ke bentuk umum integral parsial.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral f open parentheses x close parentheses times g open parentheses x close parentheses space d x end cell row blank equals cell integral u space d v end cell row blank equals cell u v minus integral v space d u end cell row blank equals cell x cubed times 1 fourth open parentheses x plus 1 close parentheses to the power of 4 minus integral 1 fourth open parentheses x plus 1 close parentheses to the power of 4 times 3 x squared d x end cell row blank equals cell 1 fourth x cubed open parentheses x plus 1 close parentheses to the power of 4 minus 3 over 4 integral open parentheses x plus 1 close parentheses to the power of 4 times x squared d x end cell row blank equals cell 1 fourth x cubed open parentheses x plus 1 close parentheses to the power of 4 minus 3 over 4 open parentheses fraction numerator 1 over denominator 4 plus 1 end fraction open parentheses x plus 1 close parentheses to the power of 4 plus 1 end exponent times fraction numerator 1 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus C close parentheses end cell row blank equals cell 1 fourth x cubed open parentheses x plus 1 close parentheses to the power of 4 minus 3 over 4 times 1 fifth open parentheses x plus 1 close parentheses to the power of 5 times 1 third x cubed plus C end cell row blank equals cell 1 fourth x cubed open parentheses x plus 1 close parentheses to the power of 4 minus 3 over 60 open parentheses x plus 1 close parentheses to the power of 5 x cubed plus C end cell row blank equals cell 1 fourth x cubed open parentheses x plus 1 close parentheses to the power of 4 minus 3 over 60 x cubed open parentheses x plus 1 close parentheses to the power of 5 plus C end cell row blank equals cell 1 fourth x cubed open parentheses x plus 1 close parentheses to the power of 4 minus 1 over 20 x cubed open parentheses x plus 1 close parentheses to the power of 5 plus C end cell end table

Maka, hasil dari integral parsial integral x cubed open parentheses x plus 1 close parentheses cubed d x adalah 1 fourth x cubed open parentheses x plus 1 close parentheses to the power of 4 minus 1 over 20 x cubed open parentheses x plus 1 close parentheses to the power of 5 plus C.

Roboguru

Selesaikan secara integrasi parsial masing-masing ekspresi berikut. 4.

Pembahasan Soal:

Roboguru

Selesaikan setiap integral tak tentu di bawah ini dengan cara Kino dan cara manipulasi aljabar-substitusi. Lalu, ubah jawaban cara manipulasi aljabar-substitusi ke bentuk jawaban cara Kino. a.

Pembahasan Soal:

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