<p>What is the antiderivative of e^(-x^2)?</p>
<p>Show your steps. Usually I'm on the answering side of these but today I'm gonna ask one. lol</p>
<p>I don't think it's possible..</p>
<p>Some antiderivatives can't be expressed in terms of elementary functions, and this is one of them. If you're wondering, the antiderivative of e^(-x^2) is defined as the error function erf(x).</p>
<p>However I think you can express it as an infinite series...</p>
<p>e^(-x^2) = 1/ ( e^(x^2))</p>
<p>therefore multiply and divide by ( e^(x^2) * 2x) and therefore the answer should be equal to :</p>
<p>ln[e^(x^2)] / (e^(x^2) * 2x) + c</p>
<p>yup it's impossible... if my Ti 89 can't do it, then NOONE can.. :D</p>
<p>It cannot be expressed in a finite number of functions, at least to my memory.</p>
<p>Yes, this is integrable.</p>
<p>It requires the use of series.</p>
<p>I have forgotten exactly how to do it this moment.</p>
<p>I will type in my solution later on.</p>
<p>But, the point is, it is integrable.</p>
<p>calculus+fun=error</p>
<p>This is the answer guys. I just did it, look above for an explanation</p>
<p>ln[e^(x^2)] / (e^(x^2) * 2x) + c</p>
<p>Human brain is more powerful than calculators - At least I believe that</p>
<p>Shubham- okaaay but show me all your steps to get it cause im not following u right now...</p>
<p>Shubham, that is not the correct answer. Your answer evaluated at a point, minus your answer evaluated at another point does not yield the same answer as numerical integration. </p>
<p>This is a function with no elementary antiderivative (the antiderivative is not an elementary function). You can learn more about it here: <a href="http://www.efunda.com/math/error_zeta/error_zeta.cfm%5B/url%5D">http://www.efunda.com/math/error_zeta/error_zeta.cfm</a></p>
<p>However, you can represent the antiderivative as an infinite series. </p>
<p>Here's a question: While this function does not have an elementary antiderivative, you CAN find the exact value of the integral of e^-x^2 from -infinity to +infinity. (This should only be attempted by people who have had multivariable calculus.)</p>
<p>What do you mean it does not yield the same answer? </p>
<p>I used the simple formulas we studied in Calculus BC and therefore you don't have to know multivariable calculus. There is only 1 variable x and e is a constant btw.</p>
<p>Ok let me try to clarify...</p>
<p>antiderivative of -<br>
u'/ u = ln [ u ] + c</p>
<p>now assume that e^2x is the u in above example</p>
<p>this implies that e^2x * 2 / e^2x is the derivative of ln [ e^2x ]
since the derivative of e ^2x= e^2x * 2 ( ie the derivative of 2x)</p>
<p>now use e^(-x^2) as u</p>
<p>u can be rewritten as 1/ (e^(x^2)) - I just used x^-1 = 1/x</p>
<p>now to find the antiderivative of this function we multiply inside the integral and divide outside the integral by the derivative of u</p>
<p>for the purpose of clarity I am going to use S as the symbol of integration</p>
<p>therefore basically we have done this : - </p>
<p>S ( e^(x^2) * 2x) / e^(x^2) dx / ( e^(x^2) * 2x)</p>
<p>now integral of the term between the S and dx = ln [ e^(x^2) ] + c</p>
<p>but since we are dividing the integral by u' therefore the complete answer = </p>
<p>ln [ e^(x^2)] / u' + c</p>
<p>where u' = e^( x^2) * 2x</p>
<p>try to understand this coz it is really hard to explain without the actual calculus symbols. If you need further clarification I can try to explain you further.</p>
<p>BTW where did you get this question?</p>
<p>In short, your answer is incorrect.</p>
<p>Here is my reason:
If ln[e^(x^2)]/[e^(x^2)*2x] is the antiderivative of e^(-x^2), then its derivative will be equal to e^(-x^2):</p>
<p>Notice that ln[e^(x^2)]=x^2
Thus, your antiderivative simplifies to x/[2<em>e^(x^2)], I cancelled the x's on bottom with the x^2.
Using the quotient rule, the derivative of that is 1</em>[2<em>e^(x^2)]/[2</em>e^(x^2)]^2 - x<em>2</em>2x<em>e^(x^2)/[2</em>e^(x^2)]^2
=1/2<em>e^(-x^2) - x^2</em>e^(-x^2), which is not e^(-x^2)</p>
<p>Where did the question come from? e^(-x^2) is a classic example of a function that has no "simple" antiderivative. It is also an expression that arises in probability theory.</p>
<p>Furthermore, many posters are correct in saying that you use a Taylor series to integrate.</p>
<p>Shubham if it was that easy, I think my 89 could have handled it.</p>
<p>Shubham:</p>
<p>"now to find the antiderivative of this function we multiply inside the integral and divide outside the integral by the derivative of u"</p>
<p>I don't think this is kosher, you can't just multiply inside an integral by (say) z and divide outside by z. For example, try doing this to find integral( x^2), using z=x.</p>
<p>integral(x^2) is not equal to (1/x). integral( x^3)</p>
<p>yeah exactly, i did the deriv. and shubham's not right.</p>
<p>Well, I discovered the solution without even doing any work :(</p>
<p>I wanted to at least do SOMETHING.</p>
<p>Anyway, on page 454 of the first volume of Apostol's calculus text, problem five has the solution to the topic creator's question.</p>
<p>It is the summation from n = 0 to infinity of (-1) ^ (n) * (x ^ (2 * n) ) / (n!) for all x.</p>
<p>I'd also like to point out that the solution to this question DOES NOT require any multivariable calculus.</p>
<p>You simply need to know how to use Taylor series.</p>
<p>fabrizio's right, tatlor series. however itd just be an approximation to e^(-x^2) and it wouldnt be the most fun thing to integrate in the world. lol, thanks for the replys guys</p>