Pertanyaan

Diketahui f ( x ) = 1 − sin 2 x + ∫ 0 2 π f ( x ) d x Titik potong sumbu - y dari kurva y = f(x) adalah

Diketahui

$f (x) = 1 - sin^{2} x + \int_{0}^{\frac{π}{2}} f (x) d x$

Titik potong sumbu - y dari kurva y = f(x) adalah

$begin mathsize 12px style open parentheses 0 comma 1 minus fraction numerator straight pi over denominator 5 minus straight pi end fraction close parentheses end style$
$begin mathsize 12px style open parentheses 0 comma space fraction numerator straight pi over denominator 2 plus straight pi end fraction close parentheses end style$
$begin mathsize 12px style open parentheses 0 comma 1 plus fraction numerator straight pi over denominator 4 plus 2 straight pi end fraction close parentheses end style$
$begin mathsize 12px style open parentheses 0 comma space fraction numerator straight pi over denominator 4 minus 2 straight pi end fraction close parentheses end style$
$begin mathsize 12px style open parentheses 0 comma 1 plus fraction numerator straight pi over denominator 4 minus 2 straight pi end fraction close parentheses end style$

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Habis dalam

Klaim

F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa , di mana , maka Ingat bahwa Kali 4 Sehingga didapat Titik potong sumbu - y :

Perhatikan bahwa , di mana , maka

Ingat bahwa

$begin mathsize 12px style integral subscript 0 superscript straight pi over 2 end superscript f open parentheses x close parentheses space d x equals integral subscript 0 superscript straight pi over 2 end superscript open parentheses 1 half cos space 2 x plus 1 half plus a close parentheses space d x a equals 1 half times 1 half sin space 2 x plus 1 half x plus a x right square bracket presubscript 0 presuperscript begin inline style straight pi over 2 end style end presuperscript a equals 1 fourth sin space 2 x plus 1 half x plus a x right square bracket presubscript 0 presuperscript begin inline style straight pi over 2 end style end presuperscript a equals open parentheses 1 fourth sin space 2 open parentheses straight pi over 2 close parentheses plus 1 half open parentheses straight pi over 2 close parentheses plus a open parentheses straight pi over 2 close parentheses close parentheses minus open parentheses 1 fourth sin space 2 open parentheses 0 close parentheses plus 1 half open parentheses 0 close parentheses plus a open parentheses 0 close parentheses close parentheses a equals 1 fourth open parentheses 0 close parentheses plus straight pi over 4 plus fraction numerator a straight pi over denominator 2 end fraction end style$

Kali 4

$begin mathsize 12px style 4 a equals straight pi plus 2 aπ 4 straight a minus 2 aπ equals straight pi straight a open parentheses 4 minus 2 straight pi close parentheses equals straight pi straight a equals fraction numerator straight pi over denominator 4 minus 2 straight pi end fraction end style$

Sehingga didapat

$begin mathsize 12px style f open parentheses x close parentheses equals 1 minus sin squared space x plus fraction numerator straight pi over denominator 4 minus 2 straight pi end fraction end style$

Titik potong sumbu - y :

$begin mathsize 12px style f open parentheses 0 close parentheses equals 1 minus sin squared space 0 plus fraction numerator straight pi over denominator 4 minus 2 straight pi end fraction f open parentheses 0 close parentheses equals 1 plus fraction numerator straight pi over denominator 4 minus 2 straight pi end fraction therefore space open parentheses 0 comma 1 plus fraction numerator straight pi over denominator 4 minus 2 straight pi end fraction close parentheses end style$