∫(0->1) √(4-x²) dx
Let x = 2siny,dx = 2cosy dy
x = 0,y = 0,x = 1,y = π/6
=> ∫(0->π/6) (2cosy)(2cosy) dy
= 2∫(0->π/6) (1+cos2y) dy
= 2(y+1/2 * sin2y)
= 2[π/6+1/2 * sin(π/3)]
= π/3 + √3/2
∫(-2->0) dx/(x²+2x+2)
= ∫(-2->0) dx/[(x+1)²+1]
= arctan(x+1)
= arctan(1) - arctan(-1)
= π/4 - (-π/4)
= π/2
∫(0->1) xe^x dx
= ∫(0->1) x de^x
= xe^x - ∫(0->1) e^x dx
= e - (e - 1)
= 1
∫(1->e) xlnx dx
= ∫(1->e) lnx d(x²/2)
= (1/2)x²lnx - (1/2)∫(1->e) x² * 1/x dx
= (1/2)(e² - 0) - (1/4)(e² - 1)
= (1+e²)/4