The area of the bounded region enclosed by the curve
\(y=3−|x−\frac{1}{2}|−|x+1| \)
and the x-axis is
\(\frac{9}{4}\)
\(\frac{45}{16}\)
\(\frac{27}{8}\)
\(\frac{63}{16}\)
The correct answer is (C) : \(\frac{27}{8}\)
\(y = \begin{cases} 2x-\frac{7}{2} & x < -1 \\ \frac{3}{2} & -1 \leq x \leq \frac{1}{2} \\ \frac{5}{2} - 2x & x > \frac{1}{2} \end{cases}\)
\(y=3−|x−\frac{1}{2}|−|x+1|\)
Area of shaded region (required area) =\(\frac{1}{2}(3+\frac{3}{2})⋅\frac{3}{2}=\frac{27}{8}\)
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
Integral calculus is the method that can be used to calculate the area between two curves that fall in between two intersecting curves. Similarly, we can use integration to find the area under two curves where we know the equation of two curves and their intersection points. In the given image, we have two functions f(x) and g(x) where we need to find the area between these two curves given in the shaded portion.
Area Between Two Curves With Respect to Y is
If f(y) and g(y) are continuous on [c, d] and g(y) < f(y) for all y in [c, d], then,