Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 67

Answer

$$ = 2\sqrt {1 + {x^2}} $$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\int_{ - x}^x {\sqrt {1 + {t^2}} dt} \cr & {\text{Property }}\int_a^b {f\left( x \right)} dx = \int_a^p {f\left( x \right)} dx + \int_p^b {f\left( x \right)} dx \cr & = \frac{d}{{dx}}\int_{ - x}^0 {\sqrt {1 + {t^2}} dt} + \frac{d}{{dx}}\int_0^x {\sqrt {1 + {t^2}} dt} \cr & {\text{Property }}\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx \cr & = - \frac{d}{{dx}}\int_0^{ - x} {\sqrt {1 + {t^2}} dt} + \frac{d}{{dx}}\int_0^x {\sqrt {1 + {t^2}} dt} \cr & {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr & A\left( x \right) = \int_a^x {f\left( p \right)dp{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( p \right)dp} = f\left( x \right) \cr & {\text{Use the chain rule for }}\frac{d}{{dx}}\int_0^{ - x} {\sqrt {1 + {t^2}} dt} \cr & then \cr & = - \sqrt {1 + {{\left( { - x} \right)}^2}} \left( { - 1} \right) + \sqrt {1 + {{\left( x \right)}^2}} \cr & {\text{simplify}} \cr & = \sqrt {1 + {x^2}} + \sqrt {1 + {x^2}} \cr & = 2\sqrt {1 + {x^2}} \cr} $$
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