Respuesta :
Answer:
(a) [tex]f(t) =\frac{8}{105}t^{\frac{7}{2}}+3\sin \left(t\right)+\frac{Ct^2}{2}+Dt+E[/tex]
(b) [tex]f(x)=-\sin \left(x\right)+3x^2+7x+7[/tex]
(c) The position of the particle is [tex]s(t) =-\cos \left(t\right)-\sin \left(t\right)+4[/tex]
(d) The position of the particle is [tex]s(t)=-\frac{5t^3}{6}+\frac{t^4}{12}+\frac{3t^2}{2}+\frac{77}{4}t[/tex]
Step-by-step explanation:
An anti-derivative of a function f(x) is a function whose derivative is equal to f(x). That is, if F′(x) = f(x), then F(x) is an anti-derivative of f(x).
If F(x) is any anti-derivative of f(x) then the most general anti-derivative of f(x) is called an indefinite integral and denoted,
[tex]\int\limit {f(x)} \, dx=F(x)+c[/tex], c is any constant.
(a) To find f(t) of [tex]f'''(t) = \sqrt(t) -3 cos(t)[/tex] you must:
1. [tex]f''(t) =\int\limits {\sqrt(t) -3 cos(t)} \, dt \\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\f''(t) =\int \sqrt{t}dt-\int \:3\cos \left(t\right)dt\\\\[/tex]
[tex]\int \sqrt{t}dt=\int \:t^{\frac{1}{2}}dt\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\int \sqrt{t}dt=\frac{2}{3}t^{\frac{3}{2}}[/tex]
[tex]\int \:3\cos \left(t\right)dt=3\cdot \int \cos \left(t\right)dt\\\\\mathrm{Use\:the\:common\:integral}:\quad \int \cos \left(t\right)dt=\sin \left(t\right)\\\\\int \:3\cos \left(t\right)dt=3\sin \left(t\right)[/tex]
[tex]\int \sqrt{t}-3\cos \left(t\right)dt=\frac{2}{3}t^{\frac{3}{2}}-3\sin \left(t\right)+C[/tex]
2. [tex]f'(t)=\int\limits {\frac{2}{3}t^{\frac{3}{2}}-3\sin \left(t\right)+C} \, dt[/tex]
[tex]\int (\frac{2}{3}t^{\frac{3}{2}}-3\sin \left(t\right)+C)dt=\int \frac{2}{3}t^{\frac{3}{2}}dt-\int \:3\sin \left(t\right)dt+\int \:Cdt\\\\\int (\frac{2}{3}t^{\frac{3}{2}}-3\sin \left(t\right)+C)dt=\frac{4}{15}t^{\frac{5}{2}}+3\cos \left(t\right)+Ct+D[/tex]
3. [tex]f(t)=\int\limits ({\frac{4}{15}t^{\frac{5}{2}}+3\cos \left(t\right)+Ct+D} )\, dt \\\\f(t) =\frac{8}{105}t^{\frac{7}{2}}+3\sin \left(t\right)+\frac{Ct^2}{2}+Dt+E[/tex]
(b) To find f(x) of [tex]f'''(x) = cos(x)[/tex] you must:
1. [tex]f''(x)=\int \left(cos\left(x\right)\right)dx=\sin \left(x\right)+C[/tex]
We use f''(0) = 6 to find the value of C
[tex]6=\sin \left(0\right)+C\\6=C[/tex]
Thus,
[tex]f''(x)=\sin \left(x\right)+6[/tex]
2. [tex]f'(x)=\int (\sin \left(x\right)+6)dx=\int (\sin \left(x\right)+6)dx=\int \sin \left(x\right)dx+\int \:6dx=-\cos \left(x\right)+6x+D[/tex]
We use f'(0) = 6 to find the value of D
[tex]6=-\cos \left(0\right)+6(0)+D\\D=7[/tex]
Thus,
[tex]f'(x)=-\cos \left(x\right)+6x+7[/tex]
3. [tex]f(x) =\int (-\cos \left(x\right)+6x+7)dx=\\\\-\int \cos \left(x\right)dx+\int \:6xdx+\int \:7dx=\\\\f(x)=-\sin \left(x\right)+3x^2+7x+E\\[/tex]
We use f(0) = 7 to find the value of E
[tex]7=-\sin \left(0\right)+3(0)^2+7(0)+E\\E=7[/tex]
Thus,
[tex]f(x)=-\sin \left(x\right)+3x^2+7x+7[/tex]
(c) You need to use the relation [tex]v(t)=\frac{ds}{dt}[/tex] to find the position of the particle.
[tex]ds=v(t)dt\\\\\int ds=\int v(t)dt\\\\s(t)=\int v(t)dt[/tex]
We know that [tex]v(t) = sin(t) -cos(t)[/tex], so
[tex]s(t)=\int (sin(t) -cos(t))dt\\\\s(t)= \int \sin \left(t\right)dt-\int \cos \left(t\right)dt\\\\s(t) =-\cos \left(t\right)-\sin \left(t\right)+C[/tex]
We use s(0) = 3 to find the value of C
[tex]3=-\cos \left(0\right)-\sin \left(0\right)+C\\C=4[/tex]
Therefore, the position of the particle is
[tex]s(t) =-\cos \left(t\right)-\sin \left(t\right)+4[/tex]
(d) You need to use the relation [tex]a(t)=\frac{dv}{dt}[/tex] to find the velocity of the particle.
[tex]dv = a(t)dt\\\\\int dv = \int a(t) dt\\\\v(t)=\int a(t) dt[/tex]
We know that [tex]a(t) = t^2-5t + 3[/tex], so
[tex]v(t)=\int (t^2-5t + 3) dt\\\\v(t)=\int \:t^2dt-\int \:5tdt+\int \:3dt\\\\v(t)=\frac{t^3}{3}-\frac{5t^2}{2}+3t+C[/tex]
Next, you need to use the relation [tex]v(t)=\frac{ds}{dt}[/tex] to find the position of the particle.
[tex]ds=v(t)dt\\\\\int ds=\int v(t)dt\\\\s(t)=\int v(t)dt[/tex]
We know that [tex]v(t)=\frac{t^3}{3}-\frac{5t^2}{2}+3t+C[/tex], so
[tex]s(t)=\int (\frac{t^3}{3}-\frac{5t^2}{2}+3t+C)dt\\\\s(t)=-\int \frac{5t^2}{2}dt+\int \frac{t^3}{3}dt+\int \:3tdt+\int \:Cdt\\\\s(t)=-\frac{5t^3}{6}+\frac{t^4}{12}+\frac{3t^2}{2}+Ct+D[/tex]
We use s(0) = 0, s(1) = 20 to find the value of C and D
[tex]0=-\frac{5(0)^3}{6}+\frac{(0)^4}{12}+\frac{3(0)^2}{2}+C(0)+D\\D=0[/tex]
[tex]20=-\frac{5(1)^3}{6}+\frac{(1)^4}{12}+\frac{3(1)^2}{2}+C(1)\\C=\frac{77}{4}[/tex]
Therefore, the position of the particle is
[tex]s(t)=-\frac{5t^3}{6}+\frac{t^4}{12}+\frac{3t^2}{2}+\frac{77}{4}t[/tex]
