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If $\int_{-2}^{5} \mathrm{f}(\mathrm{x}) \mathrm{dx}=4$ and $\int_{0}^{5}\{1+\mathrm{f}(\mathrm{x})\} \mathrm{dx}=7$, then what is
$\int_{-2}^{0} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ equal to?
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$\int_{-2}^{0} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ equal to?
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The correct answer is:
2
$\int_{-2}^{5} f(x) d x=4$ and $\int_{0}^{5}\{1+f(x)\} d x=7$
$\int_{0}^{5} f(x) d x=7-\int_{0}^{5} 1 d x=7-5=2$
$\int_{-2}^{0} f(x) d x=?$
$\int_{-2}^{0} f(x) d x+\int_{0}^{5} f(x) d x=4$
$\int_{-2}^{0} f(x) d x=4-\int_{0}^{5} f(x) d x$
$=4-2=2 . \quad[$ Using eqn. $(1)]$
$\int_{-2}^{0} f(x) d x=2$
$\int_{0}^{5} f(x) d x=7-\int_{0}^{5} 1 d x=7-5=2$
$\int_{-2}^{0} f(x) d x=?$
$\int_{-2}^{0} f(x) d x+\int_{0}^{5} f(x) d x=4$
$\int_{-2}^{0} f(x) d x=4-\int_{0}^{5} f(x) d x$
$=4-2=2 . \quad[$ Using eqn. $(1)]$
$\int_{-2}^{0} f(x) d x=2$
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