Calculer les intégrales suivantes :
I1=∫0πcost dt \displaystyle I_1=\int_{0}^{\pi} \cos t \, \text dt I1=∫0πcostdt
I2=∫0π2sin(t+π4) dt \displaystyle I_2=\int_{0}^{\frac{\pi}{2}} \sin \left(t+\dfrac{\pi}{4}\right) \, \text dt I2=∫02πsin(t+4π)dt
I3=∫01(t3+2t2+4t+1) dt \displaystyle I_3=\int_{0}^{1} (t^3 + 2t^2 + 4t + 1) \, \text dt I3=∫01(t3+2t2+4t+1)dt
I4=∫−33(12t17+2t3−t) dt \displaystyle I_4=\int_{-3}^{3} (12t^{17} + 2t^3 - t) \, \text dt I4=∫−33(12t17+2t3−t)dt
✓\checkmark✓ I1=[sint]0π=sinπ−sin0=0 \displaystyle I_1 = \left[\sin t\right]_{0}^{\pi} = \sin \pi - \sin 0 = 0 I1=[sint]0π=sinπ−sin0=0
✓\checkmark✓ I2=[−cos(t+π4)]0π2=−cos(π2+π4)+cosπ4=22+22=2\small \displaystyle I_2 = \left[-\cos\left(t+\frac{\pi}{4}\right)\right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}+\frac{\pi}{4}\right) + \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} I2=[−cos(t+4π)]02π=−cos(2π+4π)+cos4π=22+22=2
✓\checkmark✓ I3=[14t4+23t3+2t2+t]01=14+23+2+1=4712 \displaystyle I_3 = \left[\frac{1}{4}t^4+\frac{2}{3}t^3+2t^2+t\right]_{0}^{1} = \frac{1}{4} + \frac{2}{3} + 2 + 1 = \frac{47}{12} I3=[41t4+32t3+2t2+t]01=41+32+2+1=1247
✓\checkmark✓ On remarque que la fonction t↦12t17+2t3−tt \mapsto 12t^{17} + 2t^3 - tt↦12t17+2t3−t est impaire, donc I4=0 \displaystyle I_4 = 0 I4=0.