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Solve: f(x)=∫0xe−t2 dtf(x) = \int_{0}^{x} e^{-t^{2}} \, dt

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Evaluate definite integrals using the chain ruleright chevron icon

Full solution

Q. Solve: f(x)=∫0xe−t2 dtf(x) = \int_{0}^{x} e^{-t^{2}} \, dt
  1. Define Function f(x)f(x): The problem asks us to find the function f(x)f(x) which is defined by the integral of e(−t2)e^{(-t^2)} from 00 to xx. This is an example of an indefinite integral, where the upper limit of the integral is a variable rather than a constant.
  2. Set Up Integral: To begin solving the problem, we need to set up the integral of e−t2e^{-t^2} with respect to tt from 00 to xx. The integral is defined as:\newlinef(x)=∫0xe−t2 dtf(x) = \int_{0}^{x} e^{-t^2} \, dt\newlineThis integral does not have an elementary antiderivative, so we cannot express it in terms of elementary functions.
  3. Evaluate Integral: Since the integral cannot be evaluated using elementary functions, we can express the function f(x)f(x) as it is, meaning that f(x)f(x) is itself the integral from 00 to xx of e−t2 dte^{-t^2} \, dt. This is a common practice when dealing with integrals of functions that do not have simple antiderivatives.
  4. Express as Function: Therefore, the function f(x)f(x) is defined as:\newlinef(x)=∫0xe(−t2) dtf(x) = \int_{0}^{x} e^{(-t^2)} \, dt\newlineThis integral represents the area under the curve of e(−t2)e^{(-t^2)} from t=0t = 0 to t=xt = x.

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