Question

solve for x
\\(\int_{te}^{x} \frac{adx}{m} + \int_{0}^{\theta} a^2 d\theta = \frac{d}{d\theta}(a\theta)^2, a \
eq 0\\)

Explanation:

Step1: Simplify the derivative term

First, compute the derivative on the right-hand side using the power rule.
$$\frac{d}{d\theta}(a\theta)^2 = \frac{d}{d\theta}(a^2\theta^2) = 2a^2\theta$$

Step2: Evaluate the left integrals

Calculate the definite integrals one by one.
For the first integral:
$$\int_{te}^{x} \frac{a}{m} dx = \frac{a}{m}(x - te)$$
For the second integral:
$$\int_{0}^{\theta} a^2 d\theta = a^2(\theta - 0) = a^2\theta$$

Step3: Combine terms and isolate x

Substitute the results back into the original equation, then solve for $x$.
$$\frac{a}{m}(x - te) + a^2\theta = 2a^2\theta$$
Subtract $a^2\theta$ from both sides:
$$\frac{a}{m}(x - te) = a^2\theta$$
Multiply both sides by $\frac{m}{a}$ (since $a
eq 0$, this is valid):
$$x - te = am\theta$$
Add $te$ to both sides:
$$x = am\theta + te$$

Answer:

$x = am\theta + te$