b. Use the fact that $-1 \leq \sin(\theta) \leq 1$ to write a least algebraic upper bound for the absolute value of the integrand. $0 \leq \left| \frac{5 \sin(4x)}{x^2} \right| \leq 1 = g(x)$.
Added by Kristina R.
Close
Step 1
Therefore, $|\sin(4x)| \leq 1$. Then, $\left| \frac{5 \sin(4x)}{x^2} \right| = \frac{5 |\sin(4x)|}{x^2} \leq \frac{5(1)}{x^2} = \frac{5}{x^2}$. Show more…
Show all steps
Your feedback will help us improve your experience
Bahar Tehranipoor and 95 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Use the inequality $\sin x \leq x ,$ which holds for $x \geq 0 ,$ to find an upper bound for the value of $\int _ { 0 } ^ { 1 } \sin x d x . $
The Definite Integral
Definite Integrals and Antiderivatives
Use the inequality sin x ≤ x, which holds for x ≥ 0, to find an upper bound for the value of ∫₀¹ sin x dx. The upper bound for the value of ∫₀¹ sin x dx is (Type an integer or a simplified fraction.)
Akshaya R.
(a) If $f$ is continuous on $[a, b],$ show that $$\left|\int_{a}^{b} f(x) d x\right| \leqslant \int_{a}^{b}|f(x)| d x$$ $[$Hint:$-|f(x)| \leqslant f(x) \leqslant|f(x)| .]$ (b) Use the result of part (a) to show that $$\left|\int_{0}^{2 \pi} f(x) \sin 2 x d x\right| \leqslant \int_{0}^{2 \pi}|f(x)| d x$$
Integrals
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD