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Tentukan hasil integral tentu fungsi-fungsi berikut! 2.

Pertanyaan

Tentukan hasil integral tentu fungsi-fungsi berikut!

2. begin mathsize 14px style integral subscript negative 2 end subscript superscript 1 12 x cubed d x end style

Pembahasan Video:

Pembahasan Soal:

begin mathsize 14px style integral subscript negative 2 end subscript superscript 1 12 x cubed d x equals right enclose fraction numerator 12 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent end enclose subscript negative 2 end subscript superscript 1 space space space space space space space space space space space space space space space space space space space equals right enclose 3 x to the power of 4 end enclose subscript negative 2 end subscript superscript 1 space space space space space space space space space space space space space space space space space space space equals 3 left parenthesis 1 right parenthesis to the power of 4 minus 3 left parenthesis negative 2 right parenthesis to the power of 4 space space space space space space space space space space space space space space space space space space space space equals 3 minus 48 space space space space space space space space space space space space space space space space space space space equals negative 45 end style

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Terakhir diupdate 12 Juli 2021

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Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 1 superscript 3 left parenthesis 3 x squared minus 2 right parenthesis space d x end cell equals cell open square brackets x cubed minus 2 x close square brackets subscript 1 superscript 3 end cell row blank equals cell open parentheses 3 close parentheses cubed minus 2 open parentheses 3 close parentheses minus open parentheses open parentheses 1 close parentheses cubed minus 2 open parentheses 1 close parentheses close parentheses end cell row blank equals cell 27 minus 6 minus open parentheses 1 minus 2 close parentheses end cell row blank equals cell 21 minus left parenthesis negative 1 right parenthesis end cell row blank equals cell 21 plus 1 end cell row blank equals 22 end table end style 

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Roboguru

Tentukan integral tentu dari .

Pembahasan Soal:

Diketahui integral subscript 1 superscript 2 left parenthesis x squared plus 3 x right parenthesis space d x.

Ditanya integral tentu ?

Dengan menggunakan konsep sifat integral tentu dengan batas [b,a] yaitu :

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 1 superscript 2 left parenthesis x squared plus 3 x right parenthesis space d x end cell equals cell integral subscript 1 superscript 2 x squared space d x plus integral subscript 1 superscript 2 3 x space d x end cell row blank equals cell right enclose x cubed over 3 plus fraction numerator 3 x squared over denominator 2 end fraction end enclose table row 2 row 1 end table end cell row blank equals cell open parentheses 2 cubed over 3 plus fraction numerator 3 left parenthesis 2 right parenthesis squared over denominator 2 end fraction close parentheses minus open parentheses 1 cubed over 3 plus fraction numerator 3 left parenthesis 1 right parenthesis squared over denominator 2 end fraction close parentheses end cell row blank equals cell 8 over 3 plus 6 minus 1 third plus 3 over 2 end cell row blank equals cell 7 over 3 plus 3 over 2 plus 6 end cell row blank equals cell 5 over 6 plus 6 end cell row blank equals cell 6 5 over 6 end cell row blank equals cell 41 over 6 end cell end table

Jadi, integral tentu dari integral subscript 1 superscript 2 left parenthesis x squared plus 3 x right parenthesis space d x adalah 41 over 6.

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Roboguru

Diketahui Nilai dari adalah...

Pembahasan Soal:

f left parenthesis x right parenthesis space equals space 4 x cubed minus 5 x space d a n space g left parenthesis x right parenthesis equals space 2 x squared minus x  integral subscript 0 superscript 4 open square brackets f left parenthesis x right parenthesis minus g left parenthesis x right parenthesis close square brackets d x space equals integral subscript 0 superscript 4 open square brackets 4 x cubed minus 5 x minus 2 x squared plus x close square brackets d x  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 space space space equals integral subscript 0 superscript 4 open square brackets 4 x cubed minus 2 x squared minus 4 x close square brackets d x  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 space space space equals space x to the power of 4 minus 2 over 3 x cubed minus 2 x squared left enclose blank with 0 below and 4 on top end enclose  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 space space equals space open parentheses 4 to the power of 4 minus 2 over 3 left parenthesis 4 right parenthesis cubed minus 2 left parenthesis 4 right parenthesis squared close parentheses minus space left parenthesis 0 right parenthesis  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 space space equals 256 minus 128 over 3 minus 32 space equals 544 over 3  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 space space equals 181 1 third

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Roboguru

Hasil dari

Pembahasan Soal:

Rumus integral tentu adalah sebagai berikut:

integral subscript a superscript b f open parentheses x close parentheses d x equals open square brackets F open parentheses x close parentheses close square brackets subscript a superscript b equals F open parentheses b close parentheses minus F open parentheses a close parentheses 

Sehingga, hasil dari integral subscript negative 2 end subscript superscript 1 open parentheses x cubed minus 6 x squared plus 8 plus 2 close parentheses d x didapatkan:

  begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral subscript negative 2 end subscript superscript 1 open parentheses x cubed minus 6 x squared plus 8 x plus 2 close parentheses d x end cell row blank equals cell open square brackets fraction numerator 1 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent minus fraction numerator 6 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus fraction numerator 8 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent plus 2 x close square brackets subscript negative 2 end subscript superscript 1 end cell row blank equals cell open square brackets 1 fourth x to the power of 4 minus 6 over 3 x cubed plus 8 over 2 x squared plus 2 x close square brackets subscript negative 2 end subscript superscript 1 end cell row blank equals cell open square brackets 1 fourth x to the power of 4 minus 2 x cubed plus 4 x squared plus 2 x close square brackets subscript negative 2 end subscript superscript 1 end cell row blank equals cell open square brackets 1 fourth open parentheses 1 close parentheses to the power of 4 minus 2 open parentheses 1 close parentheses cubed plus 4 open parentheses 1 close parentheses squared plus 2 open parentheses 1 close parentheses close square brackets minus open square brackets 1 fourth open parentheses negative 2 close parentheses to the power of 4 minus 2 open parentheses negative 2 close parentheses cubed plus 4 open parentheses negative 2 close parentheses squared plus 2 open parentheses negative 2 close parentheses close square brackets end cell row blank equals cell open square brackets 1 fourth open parentheses 1 close parentheses minus 2 open parentheses 1 close parentheses plus 4 open parentheses 1 close parentheses plus 2 open parentheses 1 close parentheses close square brackets minus open square brackets 1 fourth open parentheses 16 close parentheses minus 2 open parentheses negative 8 close parentheses plus 4 open parentheses 4 close parentheses plus 2 open parentheses negative 2 close parentheses close square brackets end cell row blank equals cell open square brackets 1 fourth minus 2 plus 4 plus 2 close square brackets minus open square brackets 4 plus 16 plus 16 minus 4 close square brackets end cell row blank equals cell open square brackets 17 over 4 close square brackets minus open square brackets 32 close square brackets end cell row blank equals cell negative 111 over 4 end cell end table end style   

Jadi, hasil dari integral subscript negative 2 end subscript superscript 1 open parentheses x cubed minus 6 x squared plus 8 plus 2 close parentheses d x equals negative 111 over 4

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Roboguru

Misalkan A(t) menyatakan luas daerah di bawah kurva . Jika titik P(, 0) sehingga A() : A(1) = 1 : 8 , maka perbandingan luas trapesium ABPQ:DCPQ = ....

Pembahasan Soal:

begin mathsize 14px style straight A open parentheses straight x subscript 0 close parentheses equals integral subscript 0 superscript straight x subscript 0 end superscript space bx squared space dx  straight A open parentheses straight x subscript 0 close parentheses equals straight b over 3 straight x cubed open straight space close square brackets straight x subscript 0 over 0  straight A open parentheses straight x subscript 0 close parentheses equals straight b over 3 straight x subscript 0 superscript 3    straight A open parentheses 1 close parentheses equals integral subscript 0 superscript 1 space bx squared space dx  straight A open parentheses 1 close parentheses equals straight b over 3 straight x cubed open straight space close square brackets 1 over 0  straight A open parentheses 1 close parentheses equals straight b over 3    fraction numerator straight A open parentheses straight x subscript 0 close parentheses over denominator straight A open parentheses 1 close parentheses end fraction equals 1 over 8  fraction numerator straight b over 3 straight x subscript 0 superscript 3 over denominator straight b over 3 end fraction equals 1 over 8  straight x subscript 0 superscript 3 equals 1 over 8  straight x subscript 0 equals 1 half    Titik space straight A  straight x equals negative 1 comma space maka space straight y equals straight b open parentheses negative 1 close parentheses squared equals straight b    Titik space straight D  straight x equals 1 comma space maka space straight y equals straight b open parentheses 1 close parentheses squared equals straight b    Titik space straight Q  straight x equals 1 half comma straight space maka straight space straight y equals straight b open parentheses 1 half close parentheses squared equals 1 fourth straight b    straight L subscript ABPQ over straight L subscript DCPQ equals fraction numerator 1 half open parentheses AB plus PQ close parentheses BP over denominator 1 half open parentheses CD plus PQ close parentheses CP end fraction  straight L subscript ABPQ over straight L subscript DCPQ equals fraction numerator open parentheses straight b plus 1 fourth straight b close parentheses 3 over 2    over denominator open parentheses straight b plus 1 fourth straight b close parentheses 1 half straight space end fraction equals 3  straight L subscript ABPQ straight space colon straight space straight L subscript DCPQ equals 3 straight space colon 1 end style

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