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1.

Find the positive value of c  such that the area enclosed by the graph of  y = -x^2 +c^2 and the x-axis is 36 .

 

2.

Let \[f(t) = \int_0^t (x - 7)(x - 4)(x + 1)(e^x - 1) \left( \arctan (x) - \frac{\pi}{3} \right) \: dx.\]

 

Find all values of t where f(t) has a local minimum at t

 Jan 5, 2021
 #1
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Find the positive value of c such that the area enclosed by the graph of y = -x^2 +c^2 and the x-axis is 36 .

 

Hello Guest!

 

\(f(x)=y=-x^2+c^2=0\\ x^2=c^2\\ \color{blue}x=\pm\ c\)

 

\(\int _{-c}^{+c}(-x^2+c^2)dx=\color{blue}|c^2x-\frac{x^3}{3}|_{-c}^{+c}=36\)

 

\(|c^2x-\frac{x^3}{3} |_{-c}^{+c}= (c^2\cdot c-\frac{c^3}{3}) - (c^2\cdot (-c)-\frac{(-c)^3}{3}) =36\)

 

\(c^3-\frac{c^3}{3}-(-c^3-\frac{(-c)^3}{3})=36\\ c^3-\frac{c^3}{3}+c^3-\frac{c^3}{3}=36\\ 3c^3-c^3+3c^3-c^3=3\cdot 36\\ 4c^3=108\\ c=\sqrt[3]{\frac{108}{4}}=\sqrt[3]{27}\)

\(c=3\)

laugh  !

 Jan 6, 2021
edited by asinus  Jan 6, 2021
 #2
avatar+11238 
+1

Let

\(f(t) = \int_0^t (x - 7)(x - 4)(x + 1)(e^x - 1) \left( \arctan (x) - \frac{\pi}{3} \right) \: dx \) .

Find all values of t where f(t) has a local minimum at t.

 

Hello Guest!

 

All values of t, where \(f(x) = (x - 7)(x - 4)(x + 1)(e^x - 1) \left( \arctan (x) - \frac{\pi}{3} \right) \: \)

has a local minimum at x ar, ar  \(x\in \{0,\ 5.712\}\)

 Determined graphically.            

laugh  !

 Jan 6, 2021
edited by asinus  Jan 6, 2021
edited by asinus  Jan 6, 2021
 #3
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Hey, could you give the second part to me in a simplified form not a decimal? My teacher says to write our answers like that

Guest Jan 6, 2021

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