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Differentiating an accumulation function

Let F(x)=∫2xt3 dt.F(x) = \int_2^x t^3 \, dt. Use the Fundamental Theorem of Calculus (Part 1) to find F′(x)F'(x).

Hint

FTC Part 1 says that if F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt with ff continuous, then F′(x)=f(x)F'(x) = f(x). Here the upper limit is just xx itself, so no chain rule is needed — simply identify f(t)f(t) and substitute t=xt = x.

Step-by-step solution

The integrand is f(t)=t3f(t) = t^3, and the upper limit of integration is exactly xx (not a function of xx), so this is the direct case of FTC Part 1.

By FTC Part 1, F′(x)=f(x)F'(x) = f(x), so we replace tt with xx in the integrand t3t^3 to get F′(x)=x3F'(x) = x^3.

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