Y= E^x / 1 3e^x At 0 1/4

What is the area of the region enclosed by the two curves (y=e^x and y=2+3e^-x) and the y-axis?

Wolfram|Alpha: Making the world’s knowledge computabley=e^x and y=2+3e^-x) and the y-axisy=e^x= 3e(^-x)+2if F=e^x -->F= 3/F +2 =(3+2F)/FF^2-2F-3=0F= (2+(4+12)^.5 )/2= (2+4)/2 = 3F=3= e^x --> x=ln(3) -->y= e^(ln3)=3F= (2+(4-12)^.5 )/2= (-2)/2 = -1 rejected.Area= ∫ e^x dx from x=0 to x=ln(3) + ∫ 2+3e^-x dx from x=ln(3) to x=0Area= [ e^x] from x=0 to x=ln(3) + [2x-3e^-x] from x=ln(3) to x=0Area= [ e^x] from x=0 to x=ln(3) + [2x-3e^-x] from x=ln(3) to x=0Area= [ (e^0)-(e^ln3)] + [2ln3-3(e^-(ln3))-0+3e^0]Area= [ (1)-(3)] + [2ln3-3(1/3)-0+3(1)]Area= (-2)+(2ln3–1+3)Area= 2ln3= 2.197225Is it correct?when x=0 --> y=e^x=e^0 =1 --> (0,1)when x=0 --> y=3e^-x=2 =3(1)+2= 5 --> (0,5)we have triangle from three points (0,1) (0,5), (ln3, 3)Area of this triangle is =(1/2)*base* altitude= (1/2)*4*ln3=2*ln(3)so it is our answer.

What is the particular integral of (D^2+D+4) y=e^2x?

Given (D^2 + D + 4) y = e^2x. [Eq 1]To find a particular integral.Let us look for it in the form of y = A e^2x.Substituting in EQ 1, we get:(4A + 2A +4A) e^2x = e^2xOR, 10A = 1OR, A = 1/10Answer: One particular integral of the given equation is y=e^(2x) / 10

What is the next term in the arithmetic sequence x+1, 2x-3, x+5?

In an arithmetic sequence, difference between successive terms is the same:(2x-3)-(x+1) = (x+5)-(2x-3) x - 4 = -x + 8 2x = 12 x = 6x + 1 = 7 2x - 3 = 9 x + 5 = 11Next term = 13If you want to write this in terms of x, then we notice a pattern in 1st and 3rd terms: x+1, x+5. These differ by 4. So then 2nd and 4th terms should also differ by 4: 2x-3, 2x+1Next term: 2x+1Sequence: x+1, 2x-3, x+5, 2x+1, x+9, 2x+5, x+13, 2x+9, …

What is the volume of the solid formed by rotating the region inside the first quadrant enclosed by y = x^2 and y = 5x about the x-axis?

We want the volume of the solid formed by rotating the region inside the first quadrant enclosed by the lines [math]y=x^2[/math] and [math]y=5x[/math] about the X-axis.We will first determine the points of intersection of these two lines so as to obtain the limits of integration.[math]y=x^2[/math] and [math]y=5x \qquad \Rightarrow \qquad x^2=5x\qquad \Rightarrow \qquad x(x-5)=0.[/math][math]\Rightarrow \qquad x=0[/math] or [math]x=5.[/math][math]\Rightarrow \qquad[/math] The points of intersection are [math](0,0)[/math] and [math](5,25).[/math]It can be seen that in the interval [math](0,5), x^2 < 5x.[/math]So, to get the volume of the required solid we have to rotate the area between the equations [math]y=5x[/math] and [math]y=x^2[/math] about the X-axis and integrate it from [math]x=0[/math] or [math]x=5.[/math]The area of the annulus formed by such a rotation with centre as any arbitrary point [math]x[/math] on the X-axis is[math]\qquad \pi\left(25x^2\right)-\pi\left(x^4\right)=\pi\left(25x^2-x^4\right).[/math]Consider a small slice of thickness [math]dx,[/math] which we obtain from the required solid between any arbitrary point x and [math]x+dx[/math] on the X-axis.The volume of this slice of infinitesimal thickness [math]dx[/math] is [math]dV = \pi\left(25x^2-x^4\right)\,dx.[/math][math]\Rightarrow \qquad[/math] The required volume is, [math]V = \int_0^5\pi\left(25x^2-x^4\right)\,dx = \pi \left[\frac{25x^3}{3}-\frac{x^5}{5}\right]_0^5[/math][math]\qquad = \pi \left[\left(\frac{25 \times 5^3}{3}-\frac{5^5}{5}\right)-\left(0\right)\right] = 416.67 \pi.[/math]

How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?

There are actually two such tangents:Let y = mx + n be the tangent we're looking for, i.e. we have to determine m and n.By definition of tangent, the line y = mx + n is a tangent to the given curve if they interesect in exactly one point. So let's equate the two equations to calculate intersection(s):[math]y = mx + n = 2x - x^2,[/math][math]x^2 + (m-2)x + n = 0,[/math][math]x = \frac{m-2}{2}\pm\sqrt{\left(\frac{m-2}{2}\right)^2-n}.[/math]If the radicand (the "thing" under the square root) is positive, there will be two intersections. If it is negative, there will be no intersection. If it is zero, there will be one intersection, and the line will indeed be a tangent.So we get:[math]\left(\frac{m-2}{2}\right)^2-n=0,[/math][math]n=\left(\frac{m-2}{2}\right)^2.[/math]Now we can use the fact that the tangent passes through (2,9):[math]9=2m+n=2m+\left(\frac{m-2}{2}\right)^2=2m+\frac{\left(m-2\right)^2}{4},[/math][math]36=8m+4-4m+m^2,[/math][math]m^2+4m-32=0,[/math][math]m=-2\pm\sqrt{4+32}=-2\pm\sqrt{36}=-2\pm 6,[/math][math]m=4\ \text{or}\ m=-8.[/math]For m = 4, we get[math]n=\frac{(2-4)^2}{4}=\frac{4}{4}=1,[/math]and for m = -8, we get[math]n=\frac{(2-(-8))^2}{4}=\frac{100}{4}=25.[/math]The two solutions are therefore y = 4x + 1 and y = -8x + 25.