Let $F(x)=\int_{0}^{x} \sin \left(6 t^{2}\right) d t$
Find the MacLaurin polynomial of degree 7 for $F(x)$
Use this polynomial to estimate the value of $\int_{0}^{0,78} \sin \left(6 x^{2}\right) d x$.

Note: your answer to the last part needs to be correct to 9 decimal places.

Step 1 ：The MacLaurin series for a function \(f(x)\) is given by: \[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f''''(0)}{4!}x^4 + \frac{f'''''}{5!}x^5 + \frac{f''''''(0)}{6!}x^6 + \frac{f'''''''(0)}{7!}x^7 + ...\]

Step 2 ：We are given that \(F(x) = \int_{0}^{x} \sin \left(6 t^{2}\right) d t\).

Step 3 ：The derivative of \(F(x)\) is \(F'(x) = \sin(6x^2)\).

Step 4 ：The second derivative of \(F(x)\) is \(F''(x) = 12x\cos(6x^2)\).

Step 5 ：The third derivative of \(F(x)\) is \(F'''(x) = 12\cos(6x^2) - 144x^2\sin(6x^2)\).

Step 6 ：The fourth derivative of \(F(x)\) is \(F''''(x) = -288x\cos(6x^2) - 144\sin(6x^2) + 1728x^4\sin(6x^2)\).

Step 7 ：The fifth derivative of \(F(x)\) is \(F'''''(x) = 1728\sin(6x^2) + 3456x^3\cos(6x^2) - 288\cos(6x^2) - 10368x^4\sin(6x^2)\).

Step 8 ：The sixth derivative of \(F(x)\) is \(F''''''(x) = -20736x^4\cos(6x^2) + 3456x^3\sin(6x^2) + 1728\cos(6x^2) - 62208x^6\sin(6x^2)\).

Step 9 ：The seventh derivative of \(F(x)\) is \(F'''''''(x) = -124416x^6\cos(6x^2) + 20736x^4\sin(6x^2) + 62208x^5\cos(6x^2) - 103680x^7\sin(6x^2)\).

Step 10 ：Evaluating these at \(x=0\), we get: \(F(0) = 0\), \(F'(0) = 0\), \(F''(0) = 0\), \(F'''(0) = 0\), \(F''''(0) = 0\), \(F'''''(0) = 0\), \(F''''''(0) = 0\), \(F'''''''(0) = 0\).

Step 11 ：So, the MacLaurin polynomial of degree 7 for \(F(x)\) is \(0\).

Step 12 ：Therefore, the estimate of \(\int_{0}^{0.78} \sin \left(6 x^{2}\right) d x\) using this polynomial is also \(0\).

Step 13 ：However, this is a very rough estimate and the actual value of the integral is not zero. The MacLaurin series does not provide a good approximation in this case because all the derivatives of \(F(x)\) at \(x=0\) are zero.

Step 14 ：\(\boxed{0}\)