If f(x)=| x^4 x^3 1 e^x+e^-x e^x-e^-x 2 secx cosecx 3 |, then ∫-π/

step 1 So in the given question we have a function that is given as f of x is equal to the determinant

If f(x)=| x^4 x^3 1 e^x+e^-x e^x-e^-x 2 ...

If $\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\mathrm{x}^{4} & \mathrm{x}^{3} & 1 \\ \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}} & \mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}} & 2 \\ \sec \mathrm{x} & \operatorname{cosecx} & 3\end{array}\right|$, then $\int_{-\pi / 4}^{\pi / 4} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$ equals (a) 4 (b) $\frac{\pi}{2}$ (c) 0 (d) $\frac{\pi}{4}$

If $\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\mathrm{x}^{4} & \mathrm{x}^{3} & 1 \\ \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}} & \mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}} & 2 \\ \sec \mathrm{x} & \operatorname{cosecx} & 3\end{array}\right|$, then $\int_{-\pi / 4}^{\pi / 4} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$ equals
(a) 4
(b) $\frac{\pi}{2}$
(c) 0
(d) $\frac{\pi}{4}$

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Determinants

A determinant is a scalar value computed from the elements of a square matrix that encapsulates certain properties of the matrix, such as invertibility and volume scaling in linear transformations. In problems involving determinants with function entries, properties like linearity and multilinearity allow the structure of the determinant to reveal symmetries that can simplify computations.

Even and Odd Functions

Even functions satisfy f(x) = f(–x), and odd functions satisfy f(x) = –f(–x) for all x in the domain. Recognizing whether a function is even or odd is crucial because it directly impacts the evaluation of integrals over symmetric intervals, and in many cases, the function’s parity can lead to significant simplification.

Definite Integration over Symmetric Intervals

When integrating a function over an interval that is symmetric about zero, the function’s parity plays a key role. Specifically, the integral of an odd function over any symmetric interval [–a, a] is zero, while the integral of an even function may be simplified by doubling the integral from 0 to a. This property is highly useful in reducing the complexity of integral evaluations.

Transcript

Please give Ace some feedback