nilai integral

<p><strong>Soal a</strong><br>∫ dari -π sampai π e^x sin(2x) dx</p><p>Gunakan rumus integral:<br>∫ e^x sin(ax) dx = e^x (sin(ax) – a cos(ax)) / (1 + a²) + C</p><p>Untuk a = 2:<br>∫ e^x sin(2x) dx = e^x (sin(2x) – 2 cos(2x)) / 5 + C</p><p>Sekarang substitusi batas:<br>F(x) = e^x (sin(2x) – 2 cos(2x)) / 5</p><p>Hitung:<br>F(π) = e^π (sin(2π) – 2 cos(2π)) / 5 = e^π (0 – 2·1)/5 = –2e^π / 5<br>F(–π) = e^(–π) (sin(–2π) – 2 cos(–2π)) / 5 = e^(–π) (0 – 2·1)/5 = –2e^(–π) / 5</p><p>Maka integral = F(π) – F(–π)<br>= (–2e^π / 5) – (–2e^(–π) / 5)<br>= (–2e^π / 5) + (2e^(–π) / 5)<br>= (2/5)(e^(–π) – e^π)</p><p><strong>Jawaban a:</strong> (2/5)(e^(–π) – e^π)</p><p><strong>Soal b</strong><br>∫ (x³ + 2x) / (x² – x + 1) dx</p><p>Langkah 1: lakukan pembagian polinomial.<br>(x³ + 2x) ÷ (x² – x + 1)</p><ul><li>Ambil suku pertama: x³ / x² = x → kalikan: x(x² – x + 1) = x³ – x² + x<br>Kurangi: (x³ + 2x) – (x³ – x² + x) = x² + x</li><li>Ambil suku berikut: x² / x² = 1 → kalikan: 1(x² – x + 1) = x² – x + 1<br>Kurangi: (x² + x) – (x² – x + 1) = 2x – 1</li></ul><p>Jadi hasil pembagian:<br>(x³ + 2x) / (x² – x + 1) = x + 1 + (2x – 1)/(x² – x + 1)</p><p>Langkah 2: integralkan.<br>∫ (x + 1) dx + ∫ (2x – 1)/(x² – x + 1) dx</p><p>= (x² / 2 + x) + ∫ (2x – 1)/(x² – x + 1) dx</p><p>Perhatikan: turunan dari (x² – x + 1) adalah (2x – 1).<br>Jadi: ∫ (2x – 1)/(x² – x + 1) dx = ln|x² – x + 1|</p><p>Maka hasilnya:<br>x²/2 + x + ln(x² – x + 1) + C</p><p><strong>Jawaban b:</strong> (x² / 2) + x + ln(x² – x + 1) + C</p><p>👉 Jadi jawaban akhir:<br>a. (2/5)(e^(–π) – e^π)<br>b. (x²/2) + x + ln(x² – x + 1) + C</p>