
How do you integrate \[{{e}^{{{x}^{2}}}}\]from 0 to 1?
Answer
528.3k+ views
Hint: In the given question, we have been asked to integrate an expression and the upper and lower limits for that given exponential expression are given. This is the definite integral and we can see that this is the non-integratable integrand. Here \[{{e}^{{{x}^{2}}}}\]is no known function i.e. we cannot find an exact value for the given integral because it cannot be derived in terms of function.
Complete step by step solution:
We have given that,
Integration of \[{{e}^{{{x}^{2}}}}\]from 0 to 1, i.e.
\[\Rightarrow \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx\]
Hence, this is the non-integratable integrand.
We can use series approximation here. i.e.
\[{{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}\]
Thus,
\[{{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n}}}{n!}}=1+{{x}^{2}}+\dfrac{{{x}^{4}}}{2!}+\dfrac{{{x}^{6}}}{3!}\]
Integrate the given expression, we get
\[\Rightarrow \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx=\left[ x+\dfrac{1}{3}{{x}^{3}}+\dfrac{1}{5\left( 2! \right)}{{x}^{5}}+\dfrac{1}{7\left( 3! \right)}{{x}^{7}} \right]_{0}^{1}\]
Putting the limits, we get
\[\Rightarrow \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx=\dfrac{1}{42}+\dfrac{1}{10}+\dfrac{1}{3}+1\]
Solving the above expression, we get
\[\Rightarrow \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx\approx 1.48\]
Here we find the approximate value of the given integral as the given integral is the non-integratable integrand.
\[\therefore \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx\approx 1.48\]
Hence, it is the required integration.
Note: Here we need to remember that we have to put the constant term after the integration. we should know all the basic methods to integrate the given functions. There are other methods for integration. These are integration by substitution and integration by partial fractions. It is useful to know all the methods for integration so that we can choose one for computation according to the convenience and ease of calculation. We should do the calculations explicitly so that we can avoid making the errors.
Complete step by step solution:
We have given that,
Integration of \[{{e}^{{{x}^{2}}}}\]from 0 to 1, i.e.
\[\Rightarrow \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx\]
Hence, this is the non-integratable integrand.
We can use series approximation here. i.e.
\[{{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}\]
Thus,
\[{{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n}}}{n!}}=1+{{x}^{2}}+\dfrac{{{x}^{4}}}{2!}+\dfrac{{{x}^{6}}}{3!}\]
Integrate the given expression, we get
\[\Rightarrow \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx=\left[ x+\dfrac{1}{3}{{x}^{3}}+\dfrac{1}{5\left( 2! \right)}{{x}^{5}}+\dfrac{1}{7\left( 3! \right)}{{x}^{7}} \right]_{0}^{1}\]
Putting the limits, we get
\[\Rightarrow \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx=\dfrac{1}{42}+\dfrac{1}{10}+\dfrac{1}{3}+1\]
Solving the above expression, we get
\[\Rightarrow \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx\approx 1.48\]
Here we find the approximate value of the given integral as the given integral is the non-integratable integrand.
\[\therefore \int_{0}^{1}{{{e}^{{{x}^{2}}}}}dx\approx 1.48\]
Hence, it is the required integration.
Note: Here we need to remember that we have to put the constant term after the integration. we should know all the basic methods to integrate the given functions. There are other methods for integration. These are integration by substitution and integration by partial fractions. It is useful to know all the methods for integration so that we can choose one for computation according to the convenience and ease of calculation. We should do the calculations explicitly so that we can avoid making the errors.
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