1). Tentukan nilai dari integral batas atas 1 batas bawah 1 (x^2 + 2x - 3) dx . 2). Tentukan nilai p jika integral batas atas p batas bawah 1 (2x - 5) dx = 18

1.[tex] \int\limits^1_1 { x^{2} +2x-3} \, dx [/tex]
=[tex] \int\limits^1_1 { (\frac{1}{2+1} x^{2+1} ) + (2 \frac{1}{1+1} x^{1+1} ) - 3x } \, dx [/tex]
=[tex] \int\limits^1_1 { \frac{1}{3}x^{3} +x^{2} -3x } \, dx [/tex]
=[tex] (\frac{1}{3} 1^{2} + 1^{2} -3.1)-( \frac{1}{3} 1^{2} + 1^{2} -3.1)[/tex]
=0

2.[tex] \int\limits^p_1 {2x-5} \, dx [/tex]
=[tex] \int\limits^p_1 {2 \frac{1}{1+1} x^{1+1} -5x } \, dx [/tex]
=[tex] \int\limits^p_1 { x^{2} -5x} \, dx [/tex]
=[tex](p^{2} -5p) - (1^{2} -5.1)=18 [/tex]
=[tex]p^{2}-5p+4=18 [/tex]
=[tex] p^{2} -5p+4-18=0[/tex]
=[tex] p^{2} -5p-14[/tex]=0
=(p-7)(p+7)
Maka p= 7 dan -7