Approximating Definite Integrals — Question 5

Question 5 – Solution

Step 1: Determine concavity. For $f(x)=x^4$, $$f'(x)=4x^3,\qquad f''(x)=12x^2.$$ On $[0,2]$, $f''(x)\ge0$, so the graph is concave up.

Step 2: Interpret the trapezoidal approximation. For a concave-up function, each secant line lies above the graph. Therefore, $$T_n>\int_0^2x^4\,dx.$$ Step 3: Interpret the midpoint approximation. For a concave-up function, midpoint rectangles lie below the graph on average. Therefore, $$M_n<\int_0^2x^4\,dx.$$ $$\boxed{T_n>I,\qquad M_n<I}$$