Question
(a) Topic-6.6 Applying Properties of Definite Integrals
(b) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals
(c) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
(d) Topic-1.4 Estimating Limit Values from Tables
The continuous function f is defined on the closed interval −6 ≤ x ≤ 5. The figure above shows a portion of the graph of f, consisting of two line segments and a quarter of a circle centered at the point (5, 3). It is known that the point \((3,3-\sqrt{5})\) is on the graph of f.
(a) If \(\int_{-6}^{5}f(x)dx=7,\) find the value of \(\int_{-6}^{-2}f(x)dx.\) Show the work that leads to your answer
(b) Evaluate \(\int_{3}^{5}(2f'(x)+4)dx.\)
(c) The function g is given by \(g(x)=\int_{-2}^{x}f(t)dt.\) Find the absolute maximum value of g on the interval − 2 ≤ x ≤ 5. Justify your answer.
(d) Find \(\lim_{x\rightarrow 1}\frac{10^{x}-3f'(x)}{f(x)-arctan x}.\)
▶️Answer/Explanation
3(a)
\(\int_{-6}^{5} f(x) \, dx = \int_{-6}^{-2} f(x) \, dx + \int_{-2}^{5} f(x) \, dx \)
\(\implies 7 = \int_{-6}^{-2} f(x) \, dx + 2 + \left(9 – \frac{9\pi}{4}\right) \)
\(\implies \int_{-6}^{-2} f(x) \, dx = 7 – \left(11 – \frac{9\pi}{4}\right) = \frac{9\pi}{4} – 4 \)
3(b)
\(\int_{3}^{5} \big(2f'(x) + 4\big) \, dx = 2\int_{3}^{5} f'(x) \, dx + \int_{3}^{5} 4 \, dx\)
\(= 2\big(f(5) – f(3)\big) + 4(5 – 3)\)
\(= 2\big(0 – (3 – \sqrt{5})\big) + 8\)
\(= 2(-3 + \sqrt{5}) + 8 = 2 + 2\sqrt{5}\)
3(c)
\(g'(x) = f(x) = 0 \implies x = -1, \, x = \frac{1}{2}, \, x = 5\)
\(\text{On the interval } -2 \leq x \leq 5, \text{ the absolute maximum value of } g \text{ is } g(5) = 11 – \frac{9\pi}{4}.\)
3(d)
\(\lim_{x \to 1} \frac{10^x – 3f'(x)}{f(x) – \arctan x} = \frac{10^1 – 3f'(1)}{f(1) – \arctan 1}\)
\(\quad = \frac{10 – 3 \cdot 2}{1 – \arctan 1} = \frac{4}{1 – \frac{\pi}{4}}\)