Question

$\int x(1-x)^{10}\,dx =$

• A. \(\frac{(1-x)^{12}}{12}-\frac{(1-x)^{11}}{11}+C\)
• B. \(\frac{(1-x)^{12}}{12}+\frac{(1-x)^{11}}{11}+C\)
• C. \(\frac{(1-x)^{11}}{11}-\frac{(1-x)^{10}}{10}+C\)
• D. \(\frac{(1-x)^{12}}{12}+\frac{(1+x)^{11}}{11}+C\)
• E. \(\frac{(1-x)^{12}}{12}-\frac{(1+x)^{11}}{11}+C\)

Hint:

In integrals containing expressions like \(x(1-x)^n\), substitution \(u=1-x\) often makes the integral very simple. After substitution, expand and integrate term by term.

Answer 1

The Correct Option is A

Step 1: Write the integral clearly.
We need to evaluate:
\[ I=\int x(1-x)^{10}\,dx \]

Step 2: Choose a suitable substitution.
Since the expression contains \((1-x)\), take:
\[ u=1-x \] Then, \[ x=1-u \] and \[ du=-dx \quad \Rightarrow \quad dx=-du \]

Step 3: Substitute into the integral.
Now replace \(x\), \((1-x)\), and \(dx\):
\[ I=\int (1-u)u^{10}(-du) \] \[ I=-\int (1-u)u^{10}\,du \]

Step 4: Expand the integrand.
\[ (1-u)u^{10}=u^{10}-u^{11} \] So, \[ I=-\int (u^{10}-u^{11})\,du \] \[ I=\int (-u^{10}+u^{11})\,du \]

Step 5: Integrate term by term.
\[ I=-\int u^{10}\,du+\int u^{11}\,du \] \[ I=-\frac{u^{11}}{11}+\frac{u^{12}}{12}+C \]

Step 6: Substitute back \(u=1-x\).
Replacing \(u\) by \(1-x\), we get:
\[ I=-\frac{(1-x)^{11}}{11}+\frac{(1-x)^{12}}{12}+C \] Rearranging: \[ I=\frac{(1-x)^{12}}{12}-\frac{(1-x)^{11}}{11}+C \]

Step 7: Match with the given options.
Thus, \[ \boxed{\int x(1-x)^{10}\,dx=\frac{(1-x)^{12}}{12}-\frac{(1-x)^{11}}{11}+C} \] which matches option \((1)\).