We are tasked with evaluating the integral: \[ I = \int_{-10}^{10} \frac{x^{10} \sin x}{\sqrt{1 + x^{10}}} \, dx \] We observe that the integrand involves an odd function. Let’s examine the integrand for symmetry: \[ f(x) = \frac{x^{10} \sin x}{\sqrt{1 + x^{10}}} \] To check if the function is odd, we evaluate \(f(-x)\): \[ f(-x) = \frac{(-x)^{10} \sin(-x)}{\sqrt{1 + (-x)^{10}}} = \frac{x^{10} (-\sin x)}{\sqrt{1 + x^{10}}} = -\frac{x^{10} \sin x}{\sqrt{1 + x^{10}}} = -f(x) \] Since \( f(x) \) is an odd function, and we are integrating over a symmetric interval \([-10, 10]\), the integral of an odd function over a symmetric interval is zero. Therefore: \[ I = \int_{-10}^{10} f(x) \, dx = 0 \]
The correct option is (E) : \(0\)
We are given the integral:
\[ \int_{-10}^{10} \frac{x^{10} \sin x}{\sqrt{1 + x^{10}}} \, dx \]
Let us consider the **nature of the integrand**.
Define: \[ f(x) = \frac{x^{10} \sin x}{\sqrt{1 + x^{10}}} \]
Now, observe the function under \( f(-x) \):
\[ f(-x) = \frac{(-x)^{10} \sin(-x)}{\sqrt{1 + (-x)^{10}}} = \frac{x^{10} (-\sin x)}{\sqrt{1 + x^{10}}} = -\frac{x^{10} \sin x}{\sqrt{1 + x^{10}}} = -f(x) \]
Since \( f(-x) = -f(x) \), the function is odd.
And we are integrating an odd function over a symmetric interval \([-10, 10]\), so:
\[ \int_{-10}^{10} f(x) \, dx = 0 \]
Correct answer: 0
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
A stationary tank is cylindrical in shape with two hemispherical ends and is horizontal, as shown in the figure. \(R\) is the radius of the cylinder as well as of the hemispherical ends. The tank is half filled with an oil of density \(\rho\) and the rest of the space in the tank is occupied by air. The air pressure, inside the tank as well as outside it, is atmospheric. The acceleration due to gravity (\(g\)) acts vertically downward. The net horizontal force applied by the oil on the right hemispherical end (shown by the bold outline in the figure) is: