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The integral of the absolute value function, |x|, is given by the piecewise function ∫|x|dx = (x^2)/2 + C for x ≥ 0 and ∫|x|dx = (-x^2)/2 + C for x < 0, where C is the constant of integration. This is because the absolute value function changes its behavior at x = 0, resulting in two different expressions for the integral depending on the sign of x.
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How do you integrate periodic functions?
Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.
What is the integral of e raised to x cubed?
(ex)3=e3x, so int[(ex)3dx]=int[e3xdx]=e3x/3 the integral ex^3 involves a complex function useful only to integrations such as this known as the exponential integral, or En(x). The integral is:-(1/3)x*E2/3(-x3). To solve this integral, and for more information on the exponential integral, go to http://integrals.wolfram.com/index.jsp?expr=e^(x^3)&random=false
What is the integral of the derivative with respect to x of a function of x with respect to x?
∫ d/dx f(x) dx = f(x) + C C is the constant of integration.
What is the relationship of integral and differential calculus?
We say function F is an anti derivative, or indefinite integral of f if F' = f. Also, if f has an anti-derivative and is integrable on interval [a, b], then the definite integral of f from a to b is equal to F(b) - F(a) Thirdly, Let F(x) be the definite integral of integrable function f from a to x for all x in [a, b] of f, then F is an anti-derivative of f on [a,b] The definition of indefinite integral as anti-derivative, and the relation of definite integral with anti-derivative, we can conclude that integration and differentiation can be considered as two opposite operations.
Integral of e to the power of -x?
integral of e to the power -x is -e to the power -x
