Question

How do you evaluate the definite integral $\int{\left( x-{{x}^{3}} \right)dx}$ from $\left[ 0,1 \right]$?

Answer 1

In this problem we have to evaluate the given definite integral from the given interval. In this given definite integral, we have to integrate the two terms separately using basic integration. Then we can use the second fundamental theorem of calculus using the intervals, where in the first term, the upper limit is subtracted by the lower limit which is then subtracted by the second term.

Complete step-by-step solution:
We know that the given definite integral is,
$\int{\left( x-{{x}^{3}} \right)dx}$……. (1)
We also know that,
$\int{\left( a-b \right)dx=\int{\left( a \right)dx-\int{\left( b \right)dx}}}$ .
Now we can apply the above formula in (1), we get
$\Rightarrow \int{\left( x-{{x}^{3}} \right)dx=\int{\left( x \right)dx-\int{\left( {{x}^{3}} \right)dx}}}$
Now we can integrate the above step to evaluate, we get
$\Rightarrow \int{\left( x \right)dx-\int{{{\left( x \right)}^{3}}dx=\left( \dfrac{{{x}^{2}}}{2} \right)-\left( \dfrac{{{x}^{4}}}{4} \right)}}$
We also know that the given interval is $\left[ 0,1 \right]$, now we can apply this interval in the above step, we get
$\Rightarrow \int\limits_{0}^{1}{\left( x-{{x}^{3}} \right)dx=\left( \dfrac{{{x}^{2}}}{2} \right)}_{0}^{1}-\left( \dfrac{{{x}^{4}}}{4} \right)_{0}^{1}$
Now, we can apply the second fundamental theorem of calculus in the above step, we get
$\Rightarrow \left( \dfrac{{{x}^{2}}}{2} \right)_{0}^{1}-\left( \dfrac{{{x}^{4}}}{4} \right)_{0}^{1}=\dfrac{1}{2}-\dfrac{0}{2}-\left[ \dfrac{1}{4}-\dfrac{0}{4} \right]$
Now, we can subtract the above step, we get

& \Rightarrow \dfrac{1}{2}-\dfrac{1}{4} \\\ & \Rightarrow \dfrac{4-2}{8} \\\ & \Rightarrow \dfrac{2}{8}=\dfrac{1}{4} \\\ \end{aligned}$$ **Therefore, by evaluating $$\int{\left( x-{{x}^{3}} \right)dx}$$, the answer is $$\dfrac{1}{4}$$.** **Note:** Students make mistakes while integrating the terms separately. To evaluate these types of problems, we should understand the concept, formulas and properties of integration. In this problem we have used the second fundamental theorem of calculus, where in the first term, the upper limit is subtracted by the lower limit which is then subtracted by the second term. Students may also make mistakes in the cross-multiplication part which should be concentrated.