
How do you evaluate the definite integral \[\int{\left( 5x \right)dx}\] from \[\left[ 0,3 \right]\]?
Answer
545.7k+ views
Hint: To solve the given integral we will need the following properties of the integral. The first property is the integration of \[x\], we should know that \[\int{xdx=\dfrac{{{x}^{2}}}{2}+C}\], C is the constant. And the second property of integral which states that \[\int{kf\left( x \right)dx=}k\int{f\left( x \right)dx}\] there \[k\] is a constant number. We will use these properties to evaluate the given integral for the given range.
Complete step by step answer:
We are asked to evaluate the integral \[\int{\left( 5x \right)dx}\] \[\left[ 0,3 \right]\], which means that we have to find the value of \[\int\limits_{0}^{3}{5xdx}\]. We know the property of integral which states that \[\int{kf\left( x \right)dx=}k\int{f\left( x \right)dx}\] there \[k\] is a constant number. We have \[f(x)=x\], and \[k=5\]. Using this property, the above integral can be written as,
\[\Rightarrow 5\int\limits_{0}^{3}{xdx}\]
We know that the integration rule which is \[\int{xdx=\dfrac{{{x}^{2}}}{2}+C}\]. Using this rule in the above integral, the integral becomes,
\[\Rightarrow 5\int\limits_{0}^{3}{xdx}\]
\[\Rightarrow \left. \dfrac{{{x}^{2}}}{2} \right|_{0}^{3}=\dfrac{{{3}^{2}}}{2}-\dfrac{{{1}^{2}}}{2}\]
We know that the square of 3 is 9, and the square of 1 is 1 itself. Substituting these values in the above expression we get,
\[\Rightarrow \dfrac{9}{2}-\dfrac{1}{2}\]
As the denominator of both fractions is the same, we can directly subtract the numerators to calculate this subtraction. The above expression becomes,
\[\Rightarrow \dfrac{9-1}{2}=\dfrac{8}{2}\]
As the numerator and denominator have a common factor 2, we can convert the fraction to its simplest form as
\[\Rightarrow \dfrac{8}{2}=4\]
Hence, the value of integral \[\int{\left( 5x \right)dx}\] from \[\left[ 0,3 \right]\] is \[4\].
Note:
These types of questions can be solved by remembering the properties of integral, and integration of some functions. Like here we have to know the integration of \[x\], and the property of integral \[\int{kf\left( x \right)dx=}k\int{f\left( x \right)dx}\].
Complete step by step answer:
We are asked to evaluate the integral \[\int{\left( 5x \right)dx}\] \[\left[ 0,3 \right]\], which means that we have to find the value of \[\int\limits_{0}^{3}{5xdx}\]. We know the property of integral which states that \[\int{kf\left( x \right)dx=}k\int{f\left( x \right)dx}\] there \[k\] is a constant number. We have \[f(x)=x\], and \[k=5\]. Using this property, the above integral can be written as,
\[\Rightarrow 5\int\limits_{0}^{3}{xdx}\]
We know that the integration rule which is \[\int{xdx=\dfrac{{{x}^{2}}}{2}+C}\]. Using this rule in the above integral, the integral becomes,
\[\Rightarrow 5\int\limits_{0}^{3}{xdx}\]
\[\Rightarrow \left. \dfrac{{{x}^{2}}}{2} \right|_{0}^{3}=\dfrac{{{3}^{2}}}{2}-\dfrac{{{1}^{2}}}{2}\]
We know that the square of 3 is 9, and the square of 1 is 1 itself. Substituting these values in the above expression we get,
\[\Rightarrow \dfrac{9}{2}-\dfrac{1}{2}\]
As the denominator of both fractions is the same, we can directly subtract the numerators to calculate this subtraction. The above expression becomes,
\[\Rightarrow \dfrac{9-1}{2}=\dfrac{8}{2}\]
As the numerator and denominator have a common factor 2, we can convert the fraction to its simplest form as
\[\Rightarrow \dfrac{8}{2}=4\]
Hence, the value of integral \[\int{\left( 5x \right)dx}\] from \[\left[ 0,3 \right]\] is \[4\].
Note:
These types of questions can be solved by remembering the properties of integral, and integration of some functions. Like here we have to know the integration of \[x\], and the property of integral \[\int{kf\left( x \right)dx=}k\int{f\left( x \right)dx}\].
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