if f(x) = ∫₁ˣ (sin(t²) - 2) dt, then f(x) = sin(x²) 2x cos(x²) - 2 sin(x²) - 2 2x cos(x²)

Answer:

C. $\sin(x^{2}) - 2$

Explanation:

Step1: Recall the fundamental theorem of calculus

If $F(x)=\int_{a}^{x}g(t)dt$, then $F'(x) = g(x)$.

Step2: Identify the function $g(t)$

Here $g(t)=\sin(t^{2}) - 2$.

Step3: Apply the theorem

Since $f(x)=\int_{1}^{x}(\sin(t^{2}) - 2)dt$, by the fundamental - theorem of calculus, $f'(x)=\sin(x^{2}) - 2$.