Question

Let y = \(\sum_{n=1}^{3} nx^n\). The value of $\int_0^1$ y dx is .......... (decimal digits up to 2 places)
 

Hint:

When integrating a polynomial, integrate each term separately, and don't forget to apply the limits of integration.

Answer 1

Correct Answer: 1.91

Step 1: Express \( y \).
The sum \( y \) is given by: \[ y = x + 2x^2 + 3x^3. \]

Step 2: Integrate \( y \).
We need to find the integral of \( y \) from 0 to 1: \[ \int_0^1 (x + 2x^2 + 3x^3) \, dx. \] We can integrate each term separately: \[ \int_0^1 x \, dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}, \] \[ \int_0^1 2x^2 \, dx = \left[\frac{2x^3}{3}\right]_0^1 = \frac{2}{3}, \] \[ \int_0^1 3x^3 \, dx = \left[\frac{3x^4}{4}\right]_0^1 = \frac{3}{4}. \]

Step 3: Add the results.
Now, sum the results: \[ \int_0^1 y \, dx = \frac{1}{2} + \frac{2}{3} + \frac{3}{4}. \] To add these fractions, find a common denominator: \[ \int_0^1 y \, dx = \frac{6}{12} + \frac{8}{12} + \frac{9}{12} = \frac{23}{12} \approx 1.9167. \]

Step 4: Conclusion.
The value of \( \int_0^1 y \, dx \) is approximately 0.75.