Derivative of an accumulation function via FTC Part 1
Let . Find .
Hint
By the Fundamental Theorem of Calculus Part 1, where is the integrand. You don't need to compute the integral itself — just evaluate the integrand at the upper limit.
Step-by-step solution
By FTC Part 1, since is continuous on the relevant interval, the accumulation function satisfies . This means differentiating simply recovers the integrand evaluated at , with no need to find an antiderivative first.
Substituting gives . This is the exact answer, which can also be written approximately as .
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