Define \( h(x) \) in Terms of \( f(x) \):
Since \( h(x) = f(x) + |f(x)| \), we can evaluate \( h(x) \) separately on the intervals where \( f(x) \) takes different forms:
For \(-2 \leq x < 0\), \( f(x) = -x \), so \( |f(x)| = x \). Therefore:
\[ h(x) = f(x) + |f(x)| = -x + x = 0 \]
For \( 0 < x \leq 2 \), \( f(x) = x - 2 \), so \( |f(x)| = 2 - x \) (since \( x - 2 < 0 \)). Thus:
\[ h(x) = f(x) + |f(x)| = (x - 2) + (2 - x) = 0 \]
This implies \( h(x) = 0 \) on both intervals.
Calculate \( \int_{-2}^2 h(x) dx \): Since \( h(x) = 0 \) on the entire interval \(-2 \leq x \leq 2\), we have: \[ \int_{-2}^2 h(x) dx = \int_{-2}^2 0 \, dx = 0 \]
Conclusion: \( \int_{-2}^0 h(x) dx = 0 \) and \( \int_0^2 h(x) dx = 2 \).
Therefore, the answer is \(2\).
For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:
The integral is given by:
\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]
is equals to?
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32