To approximate the integral of $\sin(x^2)$ from $0$ to $0.85$ using the first three nonzero terms of its Maclaurin series, we start with the Maclaurin series expansion for $\sin(x)$:

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$

Substituting $x^2$ for $x$ gives:

$\sin(x^2) = x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} - \cdots = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} + \cdots$

Thus, the first three nonzero terms of $\sin(x^2)$ are:

$\sin(x^2) \approx x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}$

Next, we approximate the integral:

$\int_0^{0.85} \sin(x^2) \, dx \approx \int_0^{0.85} \left( x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} \right) dx$

Calculating each integral term separately:

1. First term:

$\int_0^{0.85} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{0.85} = \frac{(0.85)^3}{3} = \frac{0.614125}{3} = 0.20470833$

2. Second term:

$\int_0^{0.85} \frac{x^6}{6} \, dx = \frac{1}{6} \left[ \frac{x^7}{7} \right]_0^{0.85} = \frac{1}{6} \cdot \frac{(0.85)^7}{7}$ Calculating $(0.85)^7$:

$(0.85)^7 \approx 0.350843$ Thus,

$\int_0^{0.85} \frac{x^6}{6} \, dx \approx \frac{0.350843}{6 \cdot 7} \approx \frac{0.350843}{42} \approx 0.00834865$

3. Third term:

$\int_0^{0.85} \frac{x^{10}}{120} \, dx = \frac{1}{120} \left[ \frac{x^{11}}{11} \right]_0^{0.85} = \frac{1}{120} \cdot \frac{(0.85)^{11}}{11}$ Calculating $(0.85)^{11}$:

$(0.85)^{11} \approx 0.297677$ Thus,

$\int_0^{0.85} \frac{x^{10}}{120} \, dx \approx \frac{0.297677}{120 \cdot 11} \approx \frac{0.297677}{1320} \approx 0.00022547$

Now we sum these results to approximate the integral:

$\int_0^{0.85} \sin(x^2) \, dx \approx 0.20470833 - 0.00834865 + 0.00022547$ Calculating the total:

$\approx 0.20470833 - 0.00834865 + 0.00022547 \approx 0.19658515$

Thus, the approximate value of the integral $\int_0^{0.85} \sin(x^2) \, dx$ using the first three nonzero terms of the Maclaurin series is approximately:

$\boxed{0.19658515}$

Would you like further details or explanations? Here are some related questions you might consider:

1. How does the error in approximation relate to the number of terms used in the series?
2. Can you derive the Maclaurin series for other trigonometric functions?
3. What is the significance of the higher-order terms in the series expansion?
4. How would the approximation change if the interval was larger?
5. What numerical methods can be used to evaluate integrals more accurately?

Tip: When approximating integrals using series, consider the convergence and how many terms are necessary for the desired accuracy.