<p>Given that \(\displaystyle\int\nolimits_{-6}^{3} \mathrm{f}(x) \, \mathrm{d}x = 6\), evaluate \(\displaystyle\int\nolimits_{-6}^{-3} \mathrm{f}(x) - 1 \, \mathrm{d}x + \displaystyle\int\nolimits_{-3}^{3} \mathrm{f}(x) \, \mathrm{d}x\).</p>

Given that \(\displaystyle\displaystyle \displaystyle\int\nolimits_{-6}^{3} \mathrm{f}(x) \, \mathrm{d}x = 6\), evaluate \(\displaystyle\displaystyle \displaystyle\int\nolimits_{-6}^{-3} \mathrm{f}(x) - 1 \, \mathrm{d}x + \displaystyle\displaystyle \displaystyle\int\nolimits_{-3}^{3} \mathrm{f}(x) \, \mathrm{d}x\).

Definite Integrals and Area under graph - Understanding Results of Definite Integrals

Question 0643f067

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