Question

Find \(F^{\prime}(x)\). $$ F(x)=\int_{0}^{x^{3}} \sin t^{2} d t $$

Short answer

\(\sin(x^{6}) \cdot 3x^{2}\)

Step-by-step solution

Identify and Apply the Fundamental Theorem of Calculus

The fundamental theorem of calculus part 1 states: If \(f\) is continuous on \([a, b]\), and \(F\) is an antiderivative of \(f\) on \([a, b]\), then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). Using this, the derivative \(F^{\prime}(x)\) of the function defined by an integral \(\int_{a(x)}^{b(x)} f(t)\, dt\) is given by \(f(b(x)) \cdot b^{\prime}(x) - f(a(x)) \cdot a^{\prime}(x)\). In our case \(a(x) = 0\) and \(b(x) = x^{3}\), so \(a^{\prime}(x)=0\) and \(b^{\prime}(x)=3x^{2}\). The integrand \(f(t)\) is given by \(\sin(t^{2})\).

Calculate \(F^{\prime}(x)\)

Substitute these values into the formula from step 1: \(F^{\prime}(x) = f(b(x)) \cdot b^{\prime}(x) - f(a(x)) \cdot a^{\prime}(x) = \sin(b(x)^{2}) \cdot b^{\prime}(x) - \sin(a(x)^{2}) \cdot a^{\prime}(x) = \sin((x^{3})^{2}) \cdot 3x^{2} - 0 = \sin(x^{6}) \cdot 3x^{2}\).