dear student x2-1=-1x=0x2-1=0x=1x2-1=1x=2x2-1=2x=3x2-1=3x=2I=∫02xx2-1dx=∫01xx2-1dx+∫12xx2-1dx+∫23xx2-1dx+∫32xx2-1dx=∫01x(-1)dx+∫12x(0)dx+∫23x(1)dx+∫32x(2)dx=-1x2201+1x2223+2x2232=-12+12+4-3=1 regards

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dear student

x^{2} - 1 = - 1 x = 0 x^{2} - 1 = 0 x = 1 x^{2} - 1 = 1 x = \sqrt{2} x^{2} - 1 = 2 x = \sqrt{3} x^{2} - 1 = 3 x = 2 I = \int_{0}^{2} x [x^{2} - 1] d x = \int_{0}^{1} x [x^{2} - 1] d x + \int_{1}^{\sqrt{2}} x [x^{2} - 1] d x + \int_{\sqrt{2}}^{\sqrt{3}} x [x^{2} - 1] d x + \int_{\sqrt{3}}^{2} x [x^{2} - 1] d x = \int_{0}^{1} x (- 1) d x + \int_{1}^{\sqrt{2}} x (0) d x + \int_{\sqrt{2}}^{\sqrt{3}} x (1) d x + \int_{\sqrt{3}}^{2} x (2) d x = - 1 {[\frac{x^{2}}{2}]}_{0}^{1} + 1 {[\frac{x^{2}}{2}]}_{\sqrt{2}}^{\sqrt{3}} + 2 {[\frac{x^{2}}{2}]}_{\sqrt{3}}^{2} = - \frac{1}{2} + \frac{1}{2} + 4 - 3 = 1

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