AP Calculus BC 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation - MCQs - Exam Style Questions -
No-CalcQuestion
Evaluate \[ \lim_{n\to\infty}\sum_{k=1}^{n}\Bigg(\frac{1}{\,2+\dfrac{k}{2n}\,}\Bigg)\cdot\frac{1}{n}. \] Choose the equivalent definite integral.
(B) \(\displaystyle \int_{0}^{1/2}\frac{1}{\,2+x\,}\,dx\)
(C) \(\displaystyle \int_{0}^{1}\frac{1}{\,2+\tfrac{x}{2}\,}\,dx\)
(D) \(\displaystyle \int_{0}^{1}\frac{1}{\,2+x\,}\,dx\)
No-CalcQuestion
Which of the following limits of a Riemann sum gives the value of the definite integral \(\displaystyle \int_{-2}^{3} \!\big(x^{2}+4x\big)\,dx \)?
(B) \( \displaystyle \lim_{n\to\infty}\sum_{k=1}^{n}\frac{3}{n}\Big[\big(-2+\tfrac{3k}{n}\big)^{2}+4\!\left(-2+\tfrac{3k}{n}\right)\Big] \)
(C) \( \displaystyle \lim_{n\to\infty}\sum_{k=1}^{n}\frac{5}{n}\Big[-2+\left(\tfrac{5k}{n}\right)^{2}+4\!\left(\tfrac{5k}{n}\right)\Big] \)
(D) \( \displaystyle \lim_{n\to\infty}\sum_{k=1}^{n}\frac{5}{n}\Big[\big(-2+\tfrac{5k}{n}\big)^{2}+4\!\left(-2+\tfrac{5k}{n}\right)\Big] \)
▶️ Answer/Explanation
Correct answer: (D)
For \(\displaystyle \int_{a}^{b} f(x)\,dx\), a right–endpoint Riemann sum is \[ \lim_{n\to\infty}\sum_{k=1}^{n} f\!\big(a+k\Delta x\big)\,\Delta x,\qquad \Delta x=\frac{b-a}{n}. \] Here \(a=-2,\ b=3\Rightarrow \Delta x=\dfrac{3-(-2)}{n}=\dfrac{5}{n}\). With \(f(x)=x^{2}+4x\), \[ f\!\big(-2+\tfrac{5k}{n}\big) =\Big(-2+\tfrac{5k}{n}\Big)^{2}+4\Big(-2+\tfrac{5k}{n}\Big). \] Therefore the matching limit is \[ \boxed{ \displaystyle \lim_{n\to\infty}\sum_{k=1}^{n}\frac{5}{n} \Big[\big(-2+\tfrac{5k}{n}\big)^{2}+4\!\left(-2+\tfrac{5k}{n}\right)\Big] }. \]