Recent content by Bounceback

To clarify, this question was obtained from a list of problems involving tangents. The straight line does not need to lie on a tangent, that was my way of working the problem out. Also, my understanding of the question is that it is asking for a range for C, not one single value. To clarify why...

For what values of c is there a straight line that intersects the curve in four distinct places? x^4+c*x^3+12x^2-5x+2 I'm looking for a full answer (doesn't have to use the same method)

If I understand you correctly, you're saying to use a different method to find the value of \int_1^2 \frac {e^x}{x}\,dx, rather than differentiation under the integral sign. The question I was asked stated to use differentiation under the integral sign. If I understand you wrong, could you...

Homework Statement \int_1^2 \frac {e^x}{x}\,dx Through the use of differentiation under the integral sign. 2. The attempt at a solution Tried inputting a several times, each one resulting in another function without an elementary anti-derivative (example given below) I(a)=\int_1^2 \frac...

As an update to this, using a different method, I've a formula for any e^f(x) (assuming that the infinith derivative of any f(x) is 0, and that f(x) is differentiable) \int e^{f(x)}\,dx=e^{f(x)}*a(x) f(x)=b x=f^i(b)\ Note, f^i(f(x))=x, for\ example, if\ f(x)=x^2\ then\ f^i(x)=x^{1/2}...

To which statement of mine are you referring to? If you're referring to finding a(x) for \int e^{f(x)}\,dx=e^{f(x)}*a(x): I'm not looking for an elementary function for a(x), I'm looking for an infinite series. The series will converge if a(x) is an elementary function, otherwise the series...

There is an integral of e^{x^2} \int e^x*(a(x)+a'(x))\,dx=e^x*a(x) S=a(x)+a'(x) S-S'=a(x)-a''(x) S-S'+S''-S'''...=a(x)\pm S'''^{...} Note, S'''^{...} ends up as 0 for all S=x^constant \int e^{\sqrt{x}}\,dx=e^{\sqrt{x}}*a(x) x=\sqrt{b} dx=\frac{1}{2*\sqrt{b}}db \int \frac{e^b}{2*\sqrt{b}}\,db...

Upon imputing some more values into \int a(x)\,df(x)+a(x)=x I realized that the equations are always differing by a constant -_- Modifying the process in order to involve a constant results in: df(x)*a(x)+da(x)=d(x) \int a(x)\,df(x)+a(x)=x+C a(x)=x+C-\int a(x)\,df(x) a(x)=x+C-\int x+C-\int...

If I'm correct, an elementary function doesn't encompass an infinite series. I was intending to find something along this line for a(x). Ah, this answers the original question, thank you. Could you explain why this can't be solved (based on differential principles)? My attempts (in the...

Hmm, I'm not claiming that \int e^{f(x)}\,dx=e^{f(x)} + C. I'm claiming that \int e^{f(x)}\,dx=e^{f(x)}*a(x) + C My reasoning behind this is derivative(e^{f(x)})=e^{f(x}*a(x) Therefore the derivative of any e^(f(x)) will have a e^(f(x)) as a multiplier, and the same in the reverse for...

Just doing some testing of the two integrals I know off hand: \int e^{x^{1/3}}\,dx=3*e^{x^{1/3}}*(2-2*x^{1/3}+x^{2/3}) f'(x)*a(x)+a'(x)=1 (1/(x^{2/3}*3))*(3*(2-2*x^{1/3}+x^{2/3}))+(2*(x^1/3-1))/(x^{2/3}) Which, as it happens, equals 1 (according to the calculator) \int e^x\,dx=e^x*1...

Solve f'(x)*a(x)+a'(x)=1 For a(x) (there's another less important question at the bottom) Background behind equation (trying to find a function to integrate any e^f(x)): \int e^{f(x)}\,dx=e^{f(x)}*a(x) e^{f(x)}=e^{f(x)}*{f'(x)}*a(x)+e^{f(x)}*a'(x) 1=f'(x)*a(x)+a'(x)A few of my attempts...

iiii(a-1)!iiiiiiiiiiiiii(a-1)!iiiiiiiiiiiii(a-1)!iiiiiiiiiiiiiiiiii(a-1)! ------------ii+ii------------ii+ii------------iiiiiii+ii------------ii=ii2a-1 i(a-1)!*0!iiiiiiiiii(a-2)!*1!iiiiiiiii(a-3)!*2!iiii...iiiiiii0!*(a-1)! In case you are having trouble reading that: (a-1)!/((a-1!*0!)) +...