Cho \(\mathop \smallint \limits_{ - 3}^{ - 1} f\left( x \right)dx = 1,\mathop \smallint \limits_3^0 f\left( x \right)dx = - 2\). Tính \(\mathop \smallint \limits_0^{ - 1} f\left( x \right)dx + \mathop \smallint \limits_{ - 3}^3 f\left( x \right)dx\)
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Lời giải:
Báo saiTa có
\(\mathop \smallint \limits_{ - 3}^{ - 1} f\left( x \right)dx + \mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx + \mathop \smallint \limits_0^3 f\left( x \right)dx = \mathop \smallint \limits_{ - 3}^3 f\left( x \right)dx\)
⇒ \(\mathop \smallint \limits_{ - 3}^3 f\left( x \right)dx - \mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx = \mathop \smallint \limits_{ - 3}^3 f\left( x \right)dx + \mathop \smallint \limits_0^{ - 1} f\left( x \right)dx = \mathop \smallint \limits_{ - 3}^{ - 1} f\left( x \right)dx + \mathop \smallint \limits_0^3 f\left( x \right)dx = 1 + 2 = 3.\)