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# Calc questions

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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
+11238
+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 .

Hello Guest!

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

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

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

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

$$c=3$$

!

Jan 6, 2021
edited by asinus  Jan 6, 2021
#2
+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.

!

Jan 6, 2021
edited by asinus  Jan 6, 2021
edited by asinus  Jan 6, 2021
#3
0

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