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For the special antiderivatives involving trigonometric functions, see Trigonometric integral. [ 1 ] Generally, if the function sin x {\displaystyle \sin x} is any trigonometric function, and cos x {\displaystyle \cos x} is its derivative,
At this point we can either integrate directly, or we can first change the integrand to 2 cos 6x − 4 cos 4x + 2 cos 2x and continue from there. Either method gives Either method gives ∫ sin 2 x cos 4 x d x = − 1 24 sin 6 x + 1 8 sin 4 x − 1 8 sin 2 x + C . {\displaystyle \int \sin ^{2}x\cos 4x\,dx=-{\frac {1}{24 ...
In mathematics, the definite integral ∫ a b f ( x ) d x {\displaystyle \int _{a}^{b}f(x)\,dx} is the area of the region in the xy -plane bounded by the graph of f , the x -axis, and the lines x = a and x = b , such that area above the x -axis adds to the total, and that below the x -axis subtracts from the total.
These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.
In the mathematical field of geometric measure theory, the coarea formula expresses the integral of a function over an open set in Euclidean space in terms of integrals over the level sets of another function.
When the integrand is a constant function c, the integral is equal to the product of c and the measure of the domain of integration. If c = 1 and the domain is a subregion of R 2, the integral gives the area of the region, while if the domain is a subregion of R 3, the integral gives the volume of the region. Example. Let f(x, y) = 2 and
In this case, 1 / 3 + 1 / 5 + … + 1 / 111 < 2, but 1 / 3 + 1 / 5 + … + 1 / 113 > 2. The exact answer can be calculated using the general formula provided in the next section, and a representation of it is shown below. Fully expanded, this value turns into a fraction that involves two 2736 ...
The area of a small horizontal "slab" under the graph of f, of height dy, is equal to the measure of the slab's width times dy: ({() >}). The Lebesgue integral may then be defined by adding up the areas of these horizontal slabs. From this perspective, a key difference with the Riemann integral is that the "slabs" are no longer rectangular ...