Question

Which of the following integrals gives the length of a semicircle of radius 4, pictured above? $\int_{-4}^{4} \frac{1}{\sqrt{16 - x^2}} dx$ $\int_{-4}^{4} \sqrt{1 + (16 - x^2)^2} dx$ $\int_{-4}^{4} \frac{1}{\sqrt{16 - x^2}} dx$ $\int_{-4}^{4} \frac{8}{\sqrt{16 - x^2}} dx$ $\int_{-4}^{4} \sqrt{16 - x^2} dx$

          Which of the following integrals gives the length of a semicircle of radius 4, pictured above?
$\int_{-4}^{4} \frac{1}{\sqrt{16 - x^2}} dx$
$\int_{-4}^{4} \sqrt{1 + (16 - x^2)^2} dx$
$\int_{-4}^{4} \frac{1}{\sqrt{16 - x^2}} dx$
$\int_{-4}^{4} \frac{8}{\sqrt{16 - x^2}} dx$
$\int_{-4}^{4} \sqrt{16 - x^2} dx$

Which of the following integrals gives the length of a semicircle of radius 4, pictured above?
∫-4^4(1)/(√(16 - x^2)) dx
∫-4^4√(1 + (16 - x^2)^2) dx
∫-4^4(1)/(√(16 - x^2)) dx
∫-4^4(8)/(√(16 - x^2)) dx
∫-4^4√(16 - x^2) dx

Added by Sydney H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Elham Kordzadeh

08/23/2020

Step 1

Since we are dealing with a semicircle, we need to divide this by 2 to get the length of the semicircle. Show more…

Show all steps

Thanks for your feedback!

Which of the following integrals gives the length of a semicircle of radius 4 , pictured above? 4int_(-4)^4 (1)/(sqrt(16-x^(2)))dx int_(-4)^4 sqrt(1+(16-x^(2))^(2))dx int_(-4)^4 (1)/(sqrt(16-x^(2)))dx int_(-4)^4 (8)/(sqrt(16-x^(2)))dx pi int_(-4)^4 sqrt(16-x^(2))dx Which of the following integrals gives the length of a semicircle of radius 4, pictured above? dx /1+(16-x2)2d dx dx 2 T 16-x2d1