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what will be the integration of e^(x^2) ? solution should be with taking care of 12 th standard student what will be the integration of e^(x^2) ? solution should be with taking care of 12th standard student
2*X*e^(x^2)The usual differentiation of e^x is e^x, this is the formula we'll use here.As the term to be differentiated is e^(x^2)On differentiating it would be e^(x^2) * (differentiation of x^2)Differentiation of x^2 is 2xSo the answer is 2x*e^(x^2)Edit: I'm sry I didn't see that.Integration of E^(x^2) would be (e^(x^2))/2xThe same logic applies here the only difference is we divide the term rather than multiplying it.Formula is integral of e^xis E^xIntegral of e^(x^2) would be (e^(x^2)) divided by (differentiation of x^2)
2*X*e^(x^2)
The usual differentiation of e^x is e^x, this is the formula we'll use here.
As the term to be differentiated is e^(x^2)
On differentiating it would be e^(x^2) * (differentiation of x^2)
Differentiation of x^2 is 2x
So the answer is 2x*e^(x^2)
Edit: I'm sry I didn't see that.
Integration of E^(x^2) would be (e^(x^2))/2x
The same logic applies here the only difference is we divide the term rather than multiplying it.
Formula is integral of e^xis E^x
Integral of e^(x^2) would be (e^(x^2)) divided by (differentiation of x^2)
Dear student ere is no elementary function that describes the antiderivative of e^x^2. We can, however, express this integral in terms of an infinite series.e^x=1+x+x^2/2!+x^3/3!+x^4/4!+⋯= ∑ n=0to ∞ x^n/n!ex^2= ∑n=0 to ∞ x^2n/n!∫∑ n=0 to ∞ x^2n/n!dx = ∑n=0 to ∞ x^(2n+1) /(2n+1)n! Hence,∫e^x^2 dx = ∑n=0 to ∞ x^(2n+1)/(2n+1)n!
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