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\( \int_{-1}^{0}\left(x^{2}+1\right) \cdot d\left(x^{2}+1\right) \) |
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Answer» Let I = \(\int\limits_{-1}^0\)(x2 + 1) d (x2 + 1) Let x2 + 1 = t d(x2 + 1) = dt Limit converts from t = (-1)2 + 1 = 1 + 1 = 2 to t = 02 + 1 = 1 ∴ I = \(\int\limits_2^1tdt\) = \([\frac{t^2}2]_2^1\) = (1/2 - 1/4) = -3/2 Alternative: \(\int\limits_{-1}^0(x^2+1)d(x^2+1)\) = \([\frac{(x^2+1)^2}2]_{-1}^0\) = \((\frac12-\frac42)\) = -3/2 |
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