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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.6.61

Evaluate the integrals in Exercises 53–76.
61. ∫(from 0 to 2)dt/√(8+2t²)

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1
Identify the integral to evaluate: \(\int_0^2 \frac{dt}{\sqrt{8 + 2t^2}}\).
Factor out the constant inside the square root to simplify the integrand: rewrite \(\sqrt{8 + 2t^2}\) as \(\sqrt{2(4 + t^2)} = \sqrt{2} \sqrt{4 + t^2}\).
Rewrite the integral using this simplification: \(\int_0^2 \frac{dt}{\sqrt{2} \sqrt{4 + t^2}} = \frac{1}{\sqrt{2}} \int_0^2 \frac{dt}{\sqrt{4 + t^2}}\).
Recognize that the integral \(\int \frac{dt}{\sqrt{a^2 + t^2}}\) has a standard antiderivative: \(\ln|t + \sqrt{t^2 + a^2}| + C\), where \(a\) is a constant.
Apply the antiderivative formula with \(a = 2\), evaluate the resulting expression at the limits \(t=2\) and \(t=0\), and subtract to find the definite integral value.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals

A definite integral calculates the net area under a curve between two specified limits. It involves evaluating the integral function at the upper and lower bounds and subtracting these values. This concept is essential for solving integrals with given limits, such as from 0 to 2.
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