The correct answer is (B) : \(\frac{19}{12}\)
\(\int_{0}^{2} |2x^2 - 3x| \, dx + \int_{0}^{2} \left[x - \frac{1}{2}\right] \, dx\)
\(= \int^{3/2}_{0}(3x-2x^2)dx+\int^{2}_{3/2}(2x^2-3x)dx+\int^{1/2}_{0}-1dx+\int^{3/2}_{1/2}0dx+\int^{2}_{3/2}1dx\)
\(= (\frac{3x^2}{2}-\frac{2x^3}{3})|^{3/2}_{0}+(\frac{2x^3}{3}-\frac{3x^2}{2})|^{2}_{3/2}-\frac{1}{2}+\frac{1}{2}\)
\(\left(\frac{27}{8} - \frac{27}{12}\right) + \left(\frac{16}{3} - 6 - \frac{27}{12} + \frac{27}{8}\right)\)
\(= \frac{19}{12}\)
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
Definite integral is an operation on functions which approximates the sum of the values (of the function) weighted by the length (or measure) of the intervals for which the function takes that value.
Definite integrals - Important Formulae Handbook
A real valued function being evaluated (integrated) over the closed interval [a, b] is written as :
\(\int_{a}^{b}f(x)dx\)
Definite integrals have a lot of applications. Its main application is that it is used to find out the area under the curve of a function, as shown below:
