Integration reverses differentiation
If differentiating F(x) gives f(x), then F is an antiderivative of f. For example, differentiating x³ gives 3x², so an antiderivative of 3x² is x³. But x³ + 7 and x³ − 4 have the same derivative. Therefore the indefinite integral includes an arbitrary constant. The symbol dx identifies the variable of integration; it is not an instruction to multiply the final answer by x.
For a xⁿ with n ≠ −1, the elementary rule is ∫xⁿ, wherever the real expression is defined. Increase the by one, then divide by that new . Thus ∫x⁴, while ∫6x² dx = 2x³ + C. Check by differentiating the result. Differentiation is a strong independent check because an integration error often becomes visible immediately when the proposed answer is differentiated.
Integrate sums, roots and constants
Integration is linear: constants can be taken outside, and terms of a sum can be integrated separately. Hence. The integral of a constant k is kx, not zero. Zero is the derivative of a constant, so confusing the two operations reverses the intended rule. A single arbitrary constant suffices after combining the antiderivatives of all terms.
Rewrite roots and reciprocals as powers when appropriate. Since on the relevant real domain. Since, excluding x = 0. The −1 is a special case: the ordinary -rule denominator would become zero. Where logarithms are in scope, on intervals excluding zero. Never apply the rule by dividing by zero.
Handle simple inner linear expressions
For n ≠ −1 and constant a ≠. The additional factor a in the denominator compensates for the inner derivative. For example. Differentiating gives [3(2x + 1)² × 2]/6, which simplifies to the original integrand. Expanding the square first also works and can serve as a check.
A known condition can determine C. If and y = 5 when x = 1, integrate to obtain y = x² + 3x + C. Substitute the condition, so C = 1. The particular function is. An initial condition gives information about the original function, not about the derivative alone, so substitute it only after an antiderivative has been found.
Integrate elementary trigonometric functions
With angles expressed in radians, ∫cos and ∫sin x dx = −cos x + C. Also where defined. These follow directly from the corresponding derivatives. For a nonzero constant. A common sign error is to integrate sine as positive cosine; differentiating the proposed answer would expose it because the derivative of cosine is negative sine.
The radian convention matters in both differentiation and integration. Do not insert degree numbers into a calculus formula without a conversion or the correct degree-based factor. A definite integral from 0 to π/2 of cos x uses radian limits and equals sin(π/2) − sin 0 = 1. A trigonometric integrand with a vertical asymptote inside the interval requires more advanced treatment; elementary substitution into an antiderivative across a discontinuity is not automatically valid.
Evaluate a definite integral
A definite integral has lower and upper limits and gives an accumulated signed quantity. If F′ = f on the relevant interval, then ∫ from a to b of f(x) dx = F(b) − F(a). The final answer does not need + C because the arbitrary constants cancel. For ∫ from 1 to 3 of 2x dx, use, giving 9 − 1 = 8. Reversing the order of limits changes the sign, and identical limits give zero.
Brackets prevent substitution errors. For ∫ from 0 to 2 of (3x² − 4x) dx, an antiderivative is x³ − 2x², so the value is [8 − 8] − [0 − 0] = 0. This does not the graph encloses no geometric area. The function is negative on part of the interval and positive on another, and the signed contributions cancel. A definite integral measures signed accumulation unless the problem specifically asks for a nonnegative area.
Calculate geometric area correctly
When over [a, b], the area between its graph and the x-axis is ∫f(x) dx over those limits. If the graph lies below the axis, use the negative of its signed integral. If it crosses, find the roots inside the interval and split into pieces, taking each piece's area as nonnegative. For on [−1, 1], the signed integral is zero, while geometric area is two triangles of area 1/2 each, totaling 1.
The area between two curves is the integral of upper minus lower, splitting where their order changes. For and ² between intersections x = 0 and x = 1, the line lies above the parabola. Area is ∫ from 0 to 1 of from 0 to 1 = 1/6. Self-check: integrate 6x² + 2, evaluate ∫ from 0 to 2 of 3x² dx, and recover y from y′ = 4x with y(0) = 7. Answers are 2x³ + 2x + C, 8, and. Check units too: integrating in m/s with respect to seconds gives in metres, while total distance may require splitting where changes sign.
Recover and distinguish distance
If is from t = 0 to t = 3 s, is ∫(2t − 2) dt = [t² − 2t] from 0 to 3 = 3 m. changes sign at t = 1. The signed contribution from 0 to 1 is −1 m and from 1 to 3 is 4 m, so total distance is 1 + 4 = 5 m. This is the same distinction as signed area versus total geometric area.
Always use the original integrand to locate sign changes or discontinuities before splitting an interval. A formula for an antiderivative does not remove a forbidden point from the original function. Elementary definite-integral questions in this library use intervals on which the required antiderivatives and sign splits are valid.