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Evaluate a definite integral using an antiderivative

Note that (x2+1)0=1(x^2+1)^0 = 1 for all xx, since any nonzero quantity raised to the power 00 equals 11. Using this simplification, evaluate the definite integral:

∫01(x2+1)0⋅4x dx\int_0^1 (x^2+1)^0 \cdot 4x \, dx

Hint

First simplify the integrand using the fact that anything nonzero raised to the 0 power is 1. Then find an antiderivative F(x)F(x) of the simplified function, and apply ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a).

Step-by-step solution

Since (x2+1)0=1(x^2+1)^0 = 1 for all xx (as x2+1≠0x^2+1 \neq 0), the integrand simplifies to 4x4x. So the integral becomes ∫014x dx\int_0^1 4x\,dx. An antiderivative of 4x4x is F(x)=2x2F(x) = 2x^2, since ddx[2x2]=4x\frac{d}{dx}[2x^2] = 4x.

By FTC Part 2, ∫014x dx=F(1)−F(0)=2(1)2−2(0)2=2−0=2\int_0^1 4x\,dx = F(1) - F(0) = 2(1)^2 - 2(0)^2 = 2 - 0 = 2. Thus the value of the definite integral is 22.

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