The approximate value of int_{1}^{9} x^2 dx b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the integral $\int_{1}^{9} x^2 dx$ with $n = 4$ intervals. The width of each interval is $h = \frac{9-1}{4} = 2$. The values of $x$ are $x_0

Vedclass · Accepted Answer

$248$

Vedclass · Answer

(A) Given the integral $\int_{1}^{9} x^2 dx$ with $n = 4$ intervals. The width of each interval is $h = \frac{9-1}{4} = 2$. The values of $x$ are $x_0=1, x_1=3, x_2=5, x_3=7, x_4=9$. The corresponding values of $y = f(x) = x^2$ are: $y_0 = f(1) = 1^2 = 1$ $y_1 = f(3) = 3^2 = 9$ $y_2 = f(5) = 5^2 = 25$ $y_3 = f(7) = 7^2 = 49$ $y_4 = f(9) = 9^2 = 81$ Using the trapezoidal rule formula: $\int_{a}^{b} f(x) dx \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4]$ Substituting the values: $\int_{1}^{9} x^2 dx \approx \frac{2}{2} [1 + 2(9 + 25 + 49) + 81]$ $\int_{1}^{9} x^2 dx \approx 1 [1 + 2(83) + 81]$ $\int_{1}^{9} x^2 dx \approx 1 + 166 + 81 = 248$.

The approximate value of $\int_{1}^{9} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

Similar Questions

$\int_0^{\pi /3} \cos 3x \, dx = $

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $

$\int_{1}^{5} (|x-3| + |1-x|) dx =$

$\int_{0}^{\pi / 2} (\sin x - \cos x) \log(\sin x + \cos x) \, dx = $

$\int_0^{\pi /4} \frac{dx}{\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module