AP Calculus AB / Integration and Accumulation of Change / Question No. 35

35. \[ \frac{d}{dx} \int_{1}^{x^2} \sqrt{3 + t^2} \, dt = \]

\( \sqrt{3 + x^2} \)

\( \sqrt{3 + x^4} \)

\( 2x \sqrt{3 + x^4} \)

\( 2 \sqrt{3 + x^2} \)

Answer is: option3

\( 2x \sqrt{3 + x^4} \)

Solution:

\[ \frac{d}{dx} \int_{a}^{u(x)} f(t) \, dt = f(u(x)) \cdot \frac{du}{dx} \]

Here:

So:

\[ \frac{d}{dx} \int_{1}^{x^2} \sqrt{3 + t^2} \, dt = \sqrt{3 + (x^2)^2} \cdot \frac{d}{dx}(x^2) \]

So the derivative becomes:

\[ 2x \cdot \sqrt{3 + x^4} \]

Hence option C

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