Question

${\int }_n^{n+1}f\left(x\right)dx=\left(\begin{array}{c}n\\ n-1\end{array}\right);n\in N$, then ${\int }_1^{11}f\left(x\right)dx=$ _____

• A. $1023$
• B. $1024$
• C. $10$
• D. $55$

Answer 1

The correct option is D $55$

we can write

${\int }_1^{11}f\left(x\right)={\int }_1^{2}f\left(x\right)+{\int }_2^{3}f\left(x\right)+{\int }_3^{4}f\left(x\right)+{\int }_4^{5}f\left(x\right)+{\int }_5^{6}f\left(x\right)+{\int }_6^{7}f\left(x\right)+{\int }_7^{8}f\left(x\right)+{\int }_9^{10}f\left(x\right)+{\int }_{10}^{11}f\left(x\right)$

therefore by substituing each value we get

$1+2+3+4+5+6+7+8+9+10=55$