Answer
Verified
478.2k+ views
Hint: Integrate the given function and substitute the given values in the indefinite integration of the function. Solve them and check the parameters on which the integration depends. Use the fact that the value of \[\int{a{{x}^{n}}dx}\] is \[\dfrac{a{{x}^{n+1}}}{n+1}\]. Another way to check the dependency of the integral on the parameters is by considering the fact that the integral of odd functions from ‘-a’ to ‘a’ is zero and those of even functions is positive.
Complete step-by-step answer:
We have to calculate the value of \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}\].
We will integrate the given function and substitute the given values and then simplify it to find the value of the integral.
We know that integration of sum of functions is equal to the sum of integration of each of the functions.
Thus, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\int_{-3}^{3}{a{{x}^{5}}dx}+\int_{-3}^{3}{b{{x}^{3}}dx}+\int_{-3}^{3}{cxdx}+\int_{-3}^{3}{kdx.....\left( 1 \right)}\].
We also know that integration of function of the form \[a{{x}^{n}}\] is given by \[\int{a{{x}^{n}}dx}=\dfrac{a{{x}^{n+1}}}{n+1}\].
Thus, we have \[\int{a{{x}^{5}}dx}=\dfrac{a{{x}^{6}}}{6}\].
Similarly, we have \[\int{b{{x}^{3}}dx}=\dfrac{b{{x}^{4}}}{4}\].
We have \[\int{cxdx}=\dfrac{c{{x}^{2}}}{2}\] and \[\int{kdx=kx}\].
Substituting all the above integration in equation (1), we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\int_{-3}^{3}{a{{x}^{5}}dx}+\int_{-3}^{3}{b{{x}^{3}}dx}+\int_{-3}^{3}{cxdx}+\int_{-3}^{3}{kdx=\left[ \dfrac{a{{x}^{6}}}{6}+\dfrac{b{{x}^{4}}}{4}+\dfrac{c{{x}^{2}}}{2}+kx \right]_{x=-3}^{x=3}}\].
We will now substitute the values \[x=3\] and \[x=-3\] in the integration of the function.
Thus, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\left[ \dfrac{a{{x}^{6}}}{6}+\dfrac{b{{x}^{4}}}{4}+\dfrac{c{{x}^{2}}}{2}+kx \right]_{x=-3}^{x=3}=\dfrac{a{{\left( 3 \right)}^{6}}}{6}+\dfrac{b{{\left( 3 \right)}^{4}}}{4}+\dfrac{c{{\left( 3 \right)}^{2}}}{2}+k\left( 3 \right)-\left[ \dfrac{a{{\left( -3 \right)}^{6}}}{6}+\dfrac{b{{\left( -3 \right)}^{4}}}{4}+\dfrac{c{{\left( -3 \right)}^{2}}}{2}+k\left( -3 \right) \right]\].
Further simplifying the equation, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\dfrac{a}{6}\left( {{3}^{6}}-{{\left( -3 \right)}^{6}} \right)+\dfrac{b}{4}\left( {{3}^{4}}-{{\left( -3 \right)}^{4}} \right)+\dfrac{c}{2}\left( {{3}^{2}}-{{\left( -3 \right)}^{2}} \right)+k\left( 3-\left( -3 \right) \right)\].
Thus, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\dfrac{a}{6}\left( {{3}^{6}}-{{\left( 3 \right)}^{6}} \right)+\dfrac{b}{4}\left( {{3}^{4}}-{{\left( 3 \right)}^{4}} \right)+\dfrac{c}{2}\left( {{3}^{2}}-{{\left( 3 \right)}^{2}} \right)+k\left( 3+3 \right)\].
So, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=0+k\left( 3+3 \right)=6k\].
Hence, the value of \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}\] depends only on k, which is option (b).
Note: It’s not necessary to solve the integral completely and evaluate its value. We can also use the fact that the value of integration of odd functions from ‘-a’ to ‘a’ is zero and those of even functions is positive. Polynomials of odd degree are odd functions. Thus, the value of integration of \[a{{x}^{5}}+b{{x}^{3}}+cx\] over the given range will be zero and non – zero for \[\int_{-3}^{3}{kdx}\]. Hence, the total value of integration will be dependent on k.
Complete step-by-step answer:
We have to calculate the value of \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}\].
We will integrate the given function and substitute the given values and then simplify it to find the value of the integral.
We know that integration of sum of functions is equal to the sum of integration of each of the functions.
Thus, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\int_{-3}^{3}{a{{x}^{5}}dx}+\int_{-3}^{3}{b{{x}^{3}}dx}+\int_{-3}^{3}{cxdx}+\int_{-3}^{3}{kdx.....\left( 1 \right)}\].
We also know that integration of function of the form \[a{{x}^{n}}\] is given by \[\int{a{{x}^{n}}dx}=\dfrac{a{{x}^{n+1}}}{n+1}\].
Thus, we have \[\int{a{{x}^{5}}dx}=\dfrac{a{{x}^{6}}}{6}\].
Similarly, we have \[\int{b{{x}^{3}}dx}=\dfrac{b{{x}^{4}}}{4}\].
We have \[\int{cxdx}=\dfrac{c{{x}^{2}}}{2}\] and \[\int{kdx=kx}\].
Substituting all the above integration in equation (1), we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\int_{-3}^{3}{a{{x}^{5}}dx}+\int_{-3}^{3}{b{{x}^{3}}dx}+\int_{-3}^{3}{cxdx}+\int_{-3}^{3}{kdx=\left[ \dfrac{a{{x}^{6}}}{6}+\dfrac{b{{x}^{4}}}{4}+\dfrac{c{{x}^{2}}}{2}+kx \right]_{x=-3}^{x=3}}\].
We will now substitute the values \[x=3\] and \[x=-3\] in the integration of the function.
Thus, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\left[ \dfrac{a{{x}^{6}}}{6}+\dfrac{b{{x}^{4}}}{4}+\dfrac{c{{x}^{2}}}{2}+kx \right]_{x=-3}^{x=3}=\dfrac{a{{\left( 3 \right)}^{6}}}{6}+\dfrac{b{{\left( 3 \right)}^{4}}}{4}+\dfrac{c{{\left( 3 \right)}^{2}}}{2}+k\left( 3 \right)-\left[ \dfrac{a{{\left( -3 \right)}^{6}}}{6}+\dfrac{b{{\left( -3 \right)}^{4}}}{4}+\dfrac{c{{\left( -3 \right)}^{2}}}{2}+k\left( -3 \right) \right]\].
Further simplifying the equation, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\dfrac{a}{6}\left( {{3}^{6}}-{{\left( -3 \right)}^{6}} \right)+\dfrac{b}{4}\left( {{3}^{4}}-{{\left( -3 \right)}^{4}} \right)+\dfrac{c}{2}\left( {{3}^{2}}-{{\left( -3 \right)}^{2}} \right)+k\left( 3-\left( -3 \right) \right)\].
Thus, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\dfrac{a}{6}\left( {{3}^{6}}-{{\left( 3 \right)}^{6}} \right)+\dfrac{b}{4}\left( {{3}^{4}}-{{\left( 3 \right)}^{4}} \right)+\dfrac{c}{2}\left( {{3}^{2}}-{{\left( 3 \right)}^{2}} \right)+k\left( 3+3 \right)\].
So, we have \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=0+k\left( 3+3 \right)=6k\].
Hence, the value of \[\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}\] depends only on k, which is option (b).
Note: It’s not necessary to solve the integral completely and evaluate its value. We can also use the fact that the value of integration of odd functions from ‘-a’ to ‘a’ is zero and those of even functions is positive. Polynomials of odd degree are odd functions. Thus, the value of integration of \[a{{x}^{5}}+b{{x}^{3}}+cx\] over the given range will be zero and non – zero for \[\int_{-3}^{3}{kdx}\]. Hence, the total value of integration will be dependent on k.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE