<p>Let \(A(t)=\int_0^t\sqrt{1+x^2}\,dx\). Which are true? [MAU040]</p>
Step-by-Step Solution
Key Concept: A'(t) = \sqrt{1+t^2} > 0 \to A is strictly increasing. A(0) = \int_0^0 = 0. B is wrong (no closed form that simple).
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<p><strong>A:</strong> $A'(t)=\sqrt{1+t^2}>0$. So $A$ is strictly increasing. ✓</p>
<p><strong>B:</strong> $A(t)=\int_0^t\sqrt{1+x^2}dx\ne t\sqrt{1+t^2}$ in general. ✗ (the integral has no elementary closed form this simple)</p>
<p><strong>C:</strong> By FTC: $A'(t)=\sqrt{1+t^2}$. ✓</p>
<p><strong>D:</strong> $A(0)=\int_0^0\sqrt{1+x^2}dx=0$. ✓</p>
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Correct Answer: ['A', 'C', 'D']