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Derivative of accumulation function at x = 0.5

Let ff be the piecewise function f(t)={2tt<13t≥1f(t) = \begin{cases} 2t & t < 1 \\ 3 & t \ge 1 \end{cases} and define the accumulation function F(x)=∫0xf(t) dt.F(x) = \int_0^x f(t)\, dt.

Using the Fundamental Theorem of Calculus, find F′(0.5)F'(0.5).

Hint

By FTC Part 1, F′(x)=f(x)F'(x) = f(x) wherever ff is continuous. Determine which piece of ff applies at x=0.5x=0.5, since 0.5<10.5 < 1.

Step-by-step solution

By FTC Part 1, since ff is continuous at x=0.5x = 0.5, we have F′(x)=f(x)F'(x) = f(x) for xx near 0.50.5. Because 0.5<10.5 < 1, we use the piece f(t)=2tf(t) = 2t.

Substituting x=0.5x = 0.5 into f(x)=2xf(x) = 2x gives F′(0.5)=2(0.5)=1F'(0.5) = 2(0.5) = 1.

Derivative of accumulation function at x = 2

Using the Fundamental Theorem of Calculus, find F′(2)F'(2).

Hint

By FTC Part 1, F′(x)=f(x)F'(x) = f(x) wherever ff is continuous. Determine which piece of ff applies at x=2x=2, since 2≥12 \ge 1.

Step-by-step solution

By FTC Part 1, since ff is continuous at x=2x = 2, we have F′(x)=f(x)F'(x) = f(x) for xx near 22. Because 2≥12 \ge 1, we use the piece f(t)=3f(t) = 3.

This piece is constant, so F′(2)=3F'(2) = 3 regardless of the exact value of xx (as long as x≥1x \ge 1).

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