Réponse : Bonjour,
1)
[tex]\displaystyle \int_{-1}^{1} |x| \; dx=\int_{-1}^{0} -x \; dx+\int_{0}^{1} x \; dx=\left[-\frac{x^{2}}{2}\right]_{-1}^{0}+\left[\frac{x^{2}}{2}\right]_{0}^{1}=\frac{(-1)^{2}}{2}+\frac{1^{2}}{2}\\=\frac{1}{2}+\frac{1}{2}=1[/tex]
2)
[tex]\displaystyle \int_{\frac{\pi}{2}}^{\pi} \frac{x\cos x-\sin x}{x^{2}} \; dx=\int_{\frac{\pi}{2}}^{\pi} \frac{\cos x}{x} \; dx-\int_{\frac{\pi}{2}}^{\pi} \frac{\sin x}{x^{2}} \; dx[/tex]
On a aussi:
[tex]\displaystyle \int_{\frac{\pi}{2}}^{\pi} \frac{\sin x}{x^{2}} \; dx=\left[-\frac{1}{x} \times x\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi} -\frac{1}{x} \times \cos x \; dx=-1+\int_{\frac{\pi}{2}}^{\pi} \frac{\cos x}{x} \; dx\\[/tex]
Donc:
[tex]\displaystyle \int_{\frac{\pi}{2}}^{\pi} \frac{x \cos x-\sin x}{x^{2}} \; dx=\int_{\frac{\pi}{2}}^{\pi} \frac{\cos x}{x} \; dx-\int_{\frac{\pi}{2}}^{\pi} \frac{\sin x}{x^{2}} \; dx\\=\int_{\frac{\pi}{2}}^{\pi} \frac{\cos x}{x} \; dx+1-\int_{\frac{\pi}{2}}^{\pi} \frac{\cos x}{x} \; dx=1[/tex]
3) On a:
[tex]\displaystyle \int_{0}^{1} \frac{x}{x+2} \; dx=\int_{0}^{1} \frac{x+2-2}{x+2} \; dx=\int_{0}^{1} 1-\frac{2}{x+2} \; dx=\int_{0}^{1} 1 \; dx-\int_{0}^{1} \frac{2}{x+2} \; dx=[x]_{0}^{1}-2 \int_{0}^{1} \frac{1}{x+2} \; dx=1-2[\ln(x+2)]_{0}^{1}=1-2(\ln(1+2)-\ln(0+2))\\=1-2(\ln(3)-\ln(2))=1-2\ln(3)+2\ln(2)[/tex]
Exercice 2
1) On a:
[tex]\displaystyle \frac{a}{x-1}+\frac{b}{2-3x}=\frac{a(2-3x)+b(x-1)}{(x-1)(2-3x)}=\frac{2a-3ax+bx-b}{(x-1)(2-3x)}=\frac{(b-3a)x+2a-b}{(x-1)(2-3x)}[/tex]
Par identification, on a:
[tex]\displaystyle \left \{ {{b-3a=-10} \atop {2a-b=8}} \right. \Leftrightarrow \left \{ {{b=-10+3a} \atop {2a+10-3a=8}} \right. \Leftrightarrow \left \{ {{b=-10+3a} \atop {-a+10=8}} \right. \Leftrightarrow \left \{ {{b=-10+3 \times 2} \atop {a=2}} \right.\\ \\ \Leftrightarrow \left \{ {{b=-4} \atop {a=2}} \right.[/tex]
On a donc que:
[tex]\displaystyle \frac{8-10x}{(x-1)(2-3x)}=\frac{2}{x-1}-\frac{4}{2-3x}\\[/tex]
2) Une primitive de la fonction f est:
[tex]\displaystyle F(x)=2\ln(x-1)-4 \times -\frac{1}{3} \ln(2-3x)=2\ln(x-1)+\frac{4}{3}\ln(2-3x)[/tex]