AP Calculus BC 6.2 Approximating Areas with Riemann Sums- Exam Style Questions MCQs - New Syllabus
No-Calc Question
The definite integral \(\displaystyle \int_{0}^{4}\sqrt{x}\,dx\) is approximated by a left Riemann sum, a right Riemann sum, and a trapezoidal sum, each with 4 subintervals of equal width. If \(L\) is the value of the left Riemann sum, \(R\) is the value of the right Riemann sum, and \(T\) is the value of the trapezoidal sum, which of the following inequalities is true?
(A) \(L<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<T<R\)
(B) \(L<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<R\)
(C) \(R<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<T<L\)
(D) \(R<T<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<L\)
(B) \(L<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<R\)
(C) \(R<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<T<L\)
(D) \(R<T<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<L\)
▶️ Answer/Explanation
Detailed solution
For \(f(x)=\sqrt{x}\) on \([0,4]\):
• \(f\) is increasing \(\big(f'(x)=\tfrac{1}{2\sqrt{x}}>0\big)\) ⇒ Left sum underestimates and Right sum overestimates, so \(L<\int_{0}^{4}\!f<R\).
• \(f\) is concave down on \((0,4]\) \(\big(f”(x)=-\tfrac{1}{4x^{3/2}}<0\big)\), hence the trapezoidal sum underestimates: \(T<\int_{0}^{4}\!f\).
• With equal subintervals, \(T=\tfrac{L+R}{2}\), so \(L<T<R\). Combining yields \(L<T<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<R\).
Among the choices, the statement that must be true is the weaker inequality \(L<\displaystyle \int_{0}^{4}\sqrt{x}\,dx<R\).
✅ Answer: (B)
▶️ Answer/Explanation
Detailed solution
Subinterval \([1,\,1.5]\): midpoint \(t=1.25\), width \(\Delta t=0.5\), speed \(s(1.25)=50\) → contribution \(0.5\times 50=25\).
Subinterval \([1.5,\,3]\): midpoint \(t=2.25\), width \(\Delta t=1.5\), speed \(s(2.25)=70\) → contribution \(1.5\times 70=105\).
Subinterval \([3,\,4]\): midpoint \(t=3.5\), width \(\Delta t=1\), speed \(s(3.5)=60\) → contribution \(1\times 60=60\).
Total distance \(=25+105+60=\mathbf{190}\) miles.
✅ Correct option: B) 190