1.8 Integration

Integrate powers term by term using ∫xⁿdx=xⁿ⁺¹/(n+1)+C for n≠−1. Integration is the family of antiderivatives whose derivative is the original function.

Rewrite expressions into powers first, then integrate each term. Include C for an indefinite integral and differentiate the result to check.

∫(3x²−4x+5)dx=x³−2x²+5x+C.

The constant is not optional in an indefinite integral: different C values represent different antiderivatives.

After integrating a derivative, +C represents the unknown vertical position. A condition such as y=4 when x=1 determines C.

Integrate first, then substitute the given coordinate or initial value. In a motion problem, use the condition on displacement or velocity at the stated time.

If dy/dx=6x−2 and y(1)=5, then y=3x²−2x+C, giving C=4 and y=3x²−2x+4.

Setting C=0 assumes a particular origin that the question may not give; it is not a harmless simplification.

∫ₐᵇf(x)dx=F(b)−F(a) for an antiderivative F. It accumulates signed area: portions below the x-axis contribute negatively.

Substitute the upper and lower limits into the antiderivative, not the original integrand. Reverse limits change the sign, and equal limits give zero.

∫₀²(2x−1)dx=[x²−x]₀²=2; the graph has a small negative part but the net signed area is 2.

A definite integral is not always geometric area; use absolute values or split intervals when total area is requested.

Area between curves is ∫(upper−lower)dx or ∫(right−left)dy. Rotating a region about an axis uses washers or shells, with radius and thickness tied to the chosen variable.

Find intersections first, split where the order changes, and keep units squared or cubed. A volume formula must match the axis and region.

Rotating y=x from x=0 to 2 about the x-axis gives V=π∫₀²x²dx=8π/3.

Integrating one curve without subtracting the other can count the wrong region, and a graph crossing may require separate intervals.