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Solve the following.
For any function f(x), we have: ∫_a^b f(x) dx = ∫_a^c₁ f(x) dx + ∫_(c1)^c₂ f(x) dx + … + ∫_(cn)^b…
For any function $f(x)$, we have:
$\int_a^b f(x) \, dx = \int_a^{c_1} f(x) \, dx + \int_{c_1}^{c_2} f(x) \, dx + \dots + \int_{c_n}^b f(x) \, dx$
where $a < c_1 < c_2 < \dots < c_n < b$.
Based on above information, evaluate: $\int_0^2 |x^2 + 2x - 3| \, dx$.
Answer
Answer
Official answer keyWe have $x^2 + 2x - 3 = (x+3)(x-1)$. $|x^2+2x-3| = -(x^2+2x-3)$ if $0 < x < 1$ and $(x^2+2x-3)$ if $1 \leq x \leq 2$.
$\therefore I = \int_0^1 -(x^2+2x-3)\,dx + \int_1^2 (x^2+2x-3)\,dx$
$= -\left[\frac{x^3}{3} + \frac{2x^2}{2} - 3x\right]_0^1 + \left[\frac{x^3}{3} + \frac{2x^2}{2} - 3x\right]_1^2$
$= -\left[\left(\frac{1}{3} - 1 - 3\right) - 0\right] + \left[\left(\frac{2^3}{3} + 2^2 - 3\times2\right) - \left(\frac{1}{3} + 1 - 3\right)\right]$
$= 4$.
Final answer: 4