Question

Help please! :)
Evaluate the integral below by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry.

integral from -6 to 6 for equation (36-x^2)^(1/2) dx

Answer 1

\[\int\limits_{-6}^{6}\sqrt{36-x^2}dx\]

Answer 2

y = sqrt(36 - x^2)

y^2 = 36 - x^2

x^2 + y^2 = 36

Answer 3

This is a circle with center of (0,0) and radius 6

Answer 4

When you solve for y to get y = sqrt(36 - x^2), this is a semicircle that's above the x-axis

It's above the x-axis because the square root is positive


Compare this to

y = -sqrt(36 - x^2)

and this semicircle will be below the x-axis

Answer 5

the picture looks like this

Answer 6

does that help?

Answer 7

OMG thank you soooooo much! honestly I was completely lost. Yeah your explanation makes sense and helped me a whole lot. Thank you! :)

Answer 8

I'm glad it's finally clicking

Answer 9

if it's not to much to ask can you help me with this one as well? the problem is as follows.. If \[f(x)= \int\limits_{x}^{0}(t^3+5t^2+1)dt\] what's f"(x)? i tried differentiating it twice already and i still do not get it. Please and thank you :)

Answer 10

\[\large f(x)= \int\limits_{x}^{0}(t^3+5t^2+1)dt\]

\[\large f(x)= \frac{1}{4}t^4 + \frac{5}{3}t^3+t+C \huge|_{x}^{0}\]

\[\large f(x)= (\frac{1}{4}(0)^4 + \frac{5}{3}(0)^3+0) - (\frac{1}{4}(x)^4 + \frac{5}{3}(x)^3+x)\]

\[\large f(x)= (\frac{1}{4}(0) + \frac{5}{3}(0)+0) - (\frac{1}{4}x^4 + \frac{5}{3}x^3+x)\]

\[\large f(x)= (0 + 0+0) - (\frac{1}{4}x^4 + \frac{5}{3}x^3+x)\]

\[\large f(x)= 0 - \frac{1}{4}x^4 - \frac{5}{3}x^3-x\]

\[\large f(x)= -\frac{1}{4}x^4 - \frac{5}{3}x^3-x\]

Answer 11

\[\large f(x)= -\frac{1}{4}x^4 - \frac{5}{3}x^3-x\]

\[\large f^{\prime}(x)= \frac{d}{dx}\left[-\frac{1}{4}x^4 - \frac{5}{3}x^3-x\right]\]

\[\large f^{\prime}(x)= \frac{d}{dx}\left[-\frac{1}{4}x^4\right] - \frac{d}{dx}\left[\frac{5}{3}x^3\right]-\frac{d}{dx}\left[x\right]\]

\[\large f^{\prime}(x)= x^3 - 5x^2-1\]

Answer 12

\[\large f^{\prime}(x)= x^3 - 5x^2-1\]

\[\large f^{\prime\prime}(x)= \frac{d}{dx}\left[x^3 - 5x^2-1\right]\]

\[\large f^{\prime\prime}(x)= \frac{d}{dx}\left[x^3\right] - \left[5x^2\right]-\left[1\right]\]

\[\large f^{\prime\prime}(x)= 3x^2 -2*5x^1-0\]

\[\large f^{\prime\prime}(x)= 3x^2 -10x\]

Answer 13

oops line 3 should be this

\[\large f^{\prime\prime}(x)= \frac{d}{dx}\left[x^3\right] - \frac{d}{dx}\left[5x^2\right]-\frac{d}{dx}\left[1\right]\]

Answer 14

oh okay thank you sooo much! :)

Answer 15

np

Answer 16

uh... or you could use the fundamental theorem of calculus!$$\frac{d}{dx}\int_0^x f(x)dx=f(x)$$