Solution The midpoint rule (also known as the midpoint approximation) uses the midpoint of a subinterval for computing the height of the approximating rectangle:
∫ a b f ( x ) d x ≈ Δ x ( f ( x 0 + x 1 2 ) + f ( x 1 + x 2 2 ) + f ( x 2 + x 3 2 ) + ⋯ + f ( x n − 2 + x n − 1 2 ) + f ( x n − 1 + x n 2 ) ) \int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \Delta x \left(f{\left(\frac{x_{0} + x_{1}}{2} \right)} + f{\left(\frac{x_{1} + x_{2}}{2} \right)} + f{\left(\frac{x_{2} + x_{3}}{2} \right)}+\dots+f{\left(\frac{x_{n-2} + x_{n-1}}{2} \right)} + f{\left(\frac{x_{n-1} + x_{n}}{2} \right)}\right) a ∫ b f ( x ) d x ≈ Δ x ( f ( 2 x 0 + x 1 ) + f ( 2 x 1 + x 2 ) + f ( 2 x 2 + x 3 ) + ⋯ + f ( 2 x n − 2 + x n − 1 ) + f ( 2 x n − 1 + x n ) )
where Δ x = b − a n \Delta x = \frac{b - a}{n} Δ x = n b − a
We have that f ( x ) = sin 4 ( x ) + 7 f{\left(x \right)} = \sqrt{\sin^{4}{\left(x \right)} + 7} f ( x ) = sin 4 ( x ) + 7 a = 1 a = 1 a = 1 b = 3 b = 3 b = 3 n = 4 n = 4 n = 4
Therefore, Δ x = 3 − 1 4 = 1 2 \Delta x = \frac{3 - 1}{4} = \frac{1}{2} Δ x = 4 3 − 1 = 2 1
Divide the interval [ 1 , 3 ] \left[1, 3\right] [ 1 , 3 ] n = 4 n = 4 n = 4 Δ x = 1 2 \Delta x = \frac{1}{2} Δ x = 2 1 a = 1 a = 1 a = 1 3 2 \frac{3}{2} 2 3 2 2 2 5 2 \frac{5}{2} 2 5 3 = b 3 = b 3 = b
Now, just evaluate the function at the midpoints of the subintervals.
f ( x 0 + x 1 2 ) = f ( 1 + 3 2 2 ) = f ( 5 4 ) = sin 4 ( 5 4 ) + 7 ≈ 2.794821922941848 f{\left(\frac{x_{0} + x_{1}}{2} \right)} = f{\left(\frac{1 + \frac{3}{2}}{2} \right)} = f{\left(\frac{5}{4} \right)} = \sqrt{\sin^{4}{\left(\frac{5}{4} \right)} + 7}\approx 2.794821922941848 f ( 2 x 0 + x 1 ) = f ( 2 1 + 2 3 ) = f ( 4 5 ) = sin 4 ( 4 5 ) + 7 ≈ 2.794821922941848
f ( x 1 + x 2 2 ) = f ( 3 2 + 2 2 ) = f ( 7 4 ) = sin 4 ( 7 4 ) + 7 ≈ 2.817350905627184 f{\left(\frac{x_{1} + x_{2}}{2} \right)} = f{\left(\frac{\frac{3}{2} + 2}{2} \right)} = f{\left(\frac{7}{4} \right)} = \sqrt{\sin^{4}{\left(\frac{7}{4} \right)} + 7}\approx 2.817350905627184 f ( 2 x 1 + x 2 ) = f ( 2 2 3 + 2 ) = f ( 4 7 ) = sin 4 ( 4 7 ) + 7 ≈ 2.817350905627184
f ( x 2 + x 3 2 ) = f ( 2 + 5 2 2 ) = f ( 9 4 ) = sin 4 ( 9 4 ) + 7 ≈ 2.714130913751178 f{\left(\frac{x_{2} + x_{3}}{2} \right)} = f{\left(\frac{2 + \frac{5}{2}}{2} \right)} = f{\left(\frac{9}{4} \right)} = \sqrt{\sin^{4}{\left(\frac{9}{4} \right)} + 7}\approx 2.714130913751178 f ( 2 x 2 + x 3 ) = f ( 2 2 + 2 5 ) = f ( 4 9 ) = sin 4 ( 4 9 ) + 7 ≈ 2.714130913751178
f ( x 3 + x 4 2 ) = f ( 5 2 + 3 2 ) = f ( 11 4 ) = sin 4 ( 11 4 ) + 7 ≈ 2.649758163512828 f{\left(\frac{x_{3} + x_{4}}{2} \right)} = f{\left(\frac{\frac{5}{2} + 3}{2} \right)} = f{\left(\frac{11}{4} \right)} = \sqrt{\sin^{4}{\left(\frac{11}{4} \right)} + 7}\approx 2.649758163512828 f ( 2 x 3 + x 4 ) = f ( 2 2 5 + 3 ) = f ( 4 11 ) = sin 4 ( 4 11 ) + 7 ≈ 2.649758163512828
Finally, just sum up the above values and multiply by Δ x = 1 2 \Delta x = \frac{1}{2} Δ x = 2 1 1 2 ( 2.794821922941848 + 2.817350905627184 + 2.714130913751178 + 2.649758163512828 ) = 5.488030952916519. \frac{1}{2} \left(2.794821922941848 + 2.817350905627184 + 2.714130913751178 + 2.649758163512828\right) = 5.488030952916519. 2 1 ( 2.794821922941848 + 2.817350905627184 + 2.714130913751178 + 2.649758163512828 ) = 5.488030952916519.
