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Differentiating an accumulation function with variable limits on both ends

Let F(x)=∫x2x3t dt.F(x) = \int_{x^2}^{x^3} t \, dt.

To find F′(x)F'(x), first rewrite the integral by splitting it at a fixed point, such as 00: F(x)=∫x20t dt+∫0x3t dt=−∫0x2t dt+∫0x3t dt.F(x) = \int_{x^2}^{0} t \, dt + \int_{0}^{x^3} t \, dt = -\int_{0}^{x^2} t \, dt + \int_{0}^{x^3} t \, dt.

Using this splitting and the Fundamental Theorem of Calculus (Part 1) together with the chain rule, find F′(x)F'(x).

Hint

Apply FTC Part 1 separately to each piece: ddx∫0g(x)t dt=g(x)⋅g′(x)\frac{d}{dx}\int_0^{g(x)} t\,dt = g(x)\cdot g'(x). Don't forget the negative sign on the first piece and the chain rule factor g′(x)g'(x) for each.

Step-by-step solution

Since F(x)=−∫0x2t dt+∫0x3t dtF(x) = -\int_0^{x^2} t\,dt + \int_0^{x^3} t\,dt, differentiate each term separately using FTC Part 1 with the chain rule. For the first term, with g(x)=x2g(x) = x^2, we get ddx∫0x2t dt=x2⋅ddx(x2)=x2⋅2x=2x3\frac{d}{dx}\int_0^{x^2} t\,dt = x^2 \cdot \frac{d}{dx}(x^2) = x^2 \cdot 2x = 2x^3.

For the second term, with h(x)=x3h(x) = x^3, we get ddx∫0x3t dt=x3⋅ddx(x3)=x3⋅3x2=3x5\frac{d}{dx}\int_0^{x^3} t\,dt = x^3 \cdot \frac{d}{dx}(x^3) = x^3 \cdot 3x^2 = 3x^5. Combining both pieces with the appropriate signs gives F′(x)=−2x3+3x5F'(x) = -2x^3 + 3x^5.

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