QuestionAugust 25, 2025

Assume that sin(x) equals its Maclaurin series for all x. Use the Maclaurin series for sin(2x^2) to evaluate the integral int _(0)^0.7sin(2x^2)dx Your answer will be an infinite series. Use the first two terms to estimate its value. square

Assume that sin(x) equals its Maclaurin series for all x. Use the Maclaurin series for sin(2x^2) to evaluate the integral int _(0)^0.7sin(2x^2)dx Your answer will be an infinite series. Use the first two terms to estimate its value. square
Assume that sin(x) equals its Maclaurin series for all x. Use the Maclaurin series for sin(2x^2) to evaluate the integral int _(0)^0.7sin(2x^2)dx Your answer will be an infinite series. Use the first two terms to estimate its value. square

Solution
4.5(157 votes)

Answer

Approximately 0.228 Explanation 1. Write the Maclaurin series for \sin(2x^2) The Maclaurin series for \sin(x) is x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots. Substitute x with 2x^2: 2x^2 - \frac{(2x^2)^3}{3!} + \cdots = 2x^2 - \frac{8x^6}{6} + \cdots. 2. Integrate term by term Integrate each term separately: \int (2x^2) dx = \frac{2}{3}x^3, and \int \left(-\frac{8x^6}{6}\right) dx = -\frac{8}{42}x^7. 3. Evaluate the integral from 0 to 0.7 Substitute x = 0.7 into the integrated terms: \frac{2}{3}(0.7)^3 - \frac{8}{42}(0.7)^7.

Explanation

1. Write the Maclaurin series for $\sin(2x^2)$<br /> The Maclaurin series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$. Substitute $x$ with $2x^2$: $2x^2 - \frac{(2x^2)^3}{3!} + \cdots = 2x^2 - \frac{8x^6}{6} + \cdots$.<br />2. Integrate term by term<br /> Integrate each term separately: $\int (2x^2) dx = \frac{2}{3}x^3$, and $\int \left(-\frac{8x^6}{6}\right) dx = -\frac{8}{42}x^7$.<br />3. Evaluate the integral from 0 to 0.7<br /> Substitute $x = 0.7$ into the integrated terms: $\frac{2}{3}(0.7)^3 - \frac{8}{42}(0.7)^7$.
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