Let $f(x)= \begin{vmatrix} 1+\sin ^2 x & \cos ^2 x & \sin 2 x \\ \sin ^2 x & 1+\cos ^2 x & \sin 2 x \\ \sin ^2 x & \cos ^2 x & 1+\sin 2 x\end{vmatrix}, x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right] $ If $\alpha$ and $\beta$ respectively are the maximum and the minimum values of $f$, then
- $\beta^2+2 \sqrt{\alpha}=\frac{19}{4}$
- $\alpha^2+\beta^2=\frac{9}{2}$
- $\alpha^2-\beta^2=4 \sqrt{3}$
- $\beta^2-2 \sqrt{\alpha}=\frac{19}{4}$
The Correct Option is D
Solution and Explanation
Hence
Top Questions on integral
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
- Let $ f(x) $ be a positive function and $I_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) dx$ and $I_2 = \int_{-1}^2 f\left(x(1-x)\right) dx.$ Then the value of $\frac{I_2}{I_1}$ is equal to ____
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$x + y + z = 6$
$2x + 5y + az = 36$
$x + 2y + 3z = b$
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.