Evaluate the integral: \[ I = \int_{-3}^{3} |2 - x| dx. \]
Step 1: Identify the Breakpoint
The given function is \( |2 - x| \), which changes its definition when \( 2 - x = 0 \Rightarrow x = 2 \). Thus, we split the integral at \( x = 2 \).
Step 2: Split the Integral into Two Parts
\[ I = \int_{-3}^{2} (2 - x) dx + \int_{2}^{3} (x - 2) dx. \]
Step 3: Evaluate the First Integral
\[ \int_{-3}^{2} (2 - x) dx = \left[ 2x - \frac{x^2}{2} \right]_{-3}^{2}. \] Evaluating at \( x = 2 \): \[ 2(2) - \frac{2^2}{2} = 4 - 2 = 2. \] Evaluating at \( x = -3 \): \[ 2(-3) - \frac{(-3)^2}{2} = -6 - \frac{9}{2} = -\frac{21}{2}. \] Thus, \[ \left[ 2x - \frac{x^2}{2} \right]_{-3}^{2} = 2 - (-\frac{21}{2}) = 2 + \frac{21}{2} = \frac{25}{2}. \]
Step 4: Evaluate the Second Integral
\[ \int_{2}^{3} (x - 2) dx = \left[ \frac{x^2}{2} - 2x \right]_{2}^{3}. \] Evaluating at \( x = 3 \): \[ \frac{3^2}{2} - 2(3) = \frac{9}{2} - 6 = -\frac{3}{2}. \] Evaluating at \( x = 2 \): \[ \frac{2^2}{2} - 2(2) = \frac{4}{2} - 4 = -2. \] Thus, \[ \left[ \frac{x^2}{2} - 2x \right]_{2}^{3} = -\frac{3}{2} - (-2) = -\frac{3}{2} + 2 = \frac{1}{2}. \]
Step 5: Compute the Final Result
\[ I = \frac{25}{2} + \frac{1}{2} = \frac{26}{2} = 13. \] Thus, the final result is: \[ \boxed{13} \]
Find the area of the region (in square units) enclosed by the curves: \[ y^2 = 8(x+2), \quad y^2 = 4(1-x) \] and the Y-axis.
Evaluate the integral: \[ I = \int_{\frac{1}{\sqrt[5]{32}}}^{\frac{1}{\sqrt[5]{31}}} \frac{1}{\sqrt[5]{x^{30} + x^{25}}} dx. \]
Evaluate the integral: \[ I = \int_{-\pi}^{\pi} \frac{x \sin^3 x}{4 - \cos^2 x} dx. \]
If \[ \int \frac{3}{2\cos 3x \sqrt{2} \sin 2x} dx = \frac{3}{2} (\tan x)^{\beta} + \frac{3}{10} (\tan x)^4 + C \] then \( A = \) ?
If \[ \int \frac{2}{1+\sin x} dx = 2 \log |A(x) - B(x)| + C \] and \( 0 \leq x \leq \frac{\pi}{2} \), then \( B(\pi/4) = \) ?