Let [math: F(x)=∫03xet2 dt]\displaystyle F(x)=\int _0^{3x}{e^{t^2} dt}. Compute [math: F′(0.5)]F’(0.5). (Correct the answer to 2 decimal places.)

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We start by noting F(x) is defined as an integral with an upper limit that depends on x: F(x) = ∫ from 0 to 3x of e^{t^2} dt.
Using the Fundamental Theorem of Calculus together with the chain rule, the derivative is F'(x) = e^{(3x)^2} · d/dx(3x) = 3 e^{9x^2}.
Next, plug in x = 0.5: F'(0.5) = 3 e^{9(0.5)^2} = 3 e^{9·0.25} = 3 e^{2.25}.
Compute e^{2.25} approximately: e^{2.25} ≈ 9.4877. Therefore F'(0.5) ≈ 3 × 9.4877 ≈ 28.4631.
Rounding to two decimal places gives 28.46. This matches the provided correct result. 
Note: if you tried to differentiate under the integral sign by treating the upper limit as a function of x, you would still obtain the same chain-rule factor of 3 multiplying e^{(3x)^2}.

_MATH1013_1ABCD_2025 Subsection 5.1 (closed on 8 Nov)

Let [math: F(x)=∫03xet2 dt]\displaystyle F(x)=\int _0^{3x}{e^{t^2} dt}. Compute [math: F′(0.5)]F’(0.5). (Correct the answer to 2 decimal places.)

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