## Calculus: Early Transcendentals (2nd Edition)

$$- \sqrt {{x^4} + 1}$$
\eqalign{ & \frac{d}{{dx}}\int_x^1 {\sqrt {{t^4} + 1} dt} \cr & {\text{Property }}\int_a^b {f\left( u \right)} du = - \int_b^a {f\left( u \right)} du \cr & = - \frac{d}{{dx}}\int_1^x {\sqrt {{t^4} + 1} } dt \cr & {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr & A\left( x \right) = \int_a^x {f\left( t \right)dt{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( t \right)dt} = f\left( x \right) \cr & {\text{In this exercise}} \cr & f\left( t \right) = \sqrt {{t^4} + 1} ,{\text{ and }}a = 1 \cr & then \cr & - \frac{d}{{dx}}\int_1^x {\sqrt {{t^4} + 1} } dt = - \sqrt {{{\left( x \right)}^4} + 1} \cr & = - \sqrt {{x^4} + 1} \cr}
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