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Step-by-Step Solution
Step 1: Identify the slope of the normal
The given line is
x + 3y = -5.
Rewriting in slope-intercept form:
3y = -x - 5 \implies y = -\frac{x}{3} - \frac{5}{3}.
Thus, its slope is -\frac{1}{3}.
Since the normal to the curve at the point (a,b) is parallel to this line,
the slope of the normal, m_{\text{normal}}, is also -\frac{1}{3}.
Step 2: Relate slope of normal to slope of tangent
The slope of the tangent m_{\text{tangent}} and the slope of the normal m_{\text{normal}} at a point on a curve satisfy:
m_{\text{tangent}} \times m_{\text{normal}} = -1.
Given m_{\text{normal}} = -\frac{1}{3}, we find
m_{\text{tangent}} \times \left(-\frac{1}{3}\right) = -1
\quad \Rightarrow \quad
m_{\text{tangent}} = 3.
Step 3: Differentiate the given curve
We have
y(x) = \int_0^x \bigl(2t^2 - 15t + 10\bigr)\,dt.
By the Fundamental Theorem of Calculus, the derivative is
y'(x) = 2x^2 - 15x + 10.
Since m_{\text{tangent}} = y'(x), at x = a we must have
2a^2 - 15a + 10 = 3.
Step 4: Solve for a
Setting up the equation:
2a^2 - 15a + 10 = 3
\quad \Rightarrow \quad
2a^2 - 15a + 7 = 0.
Factorizing,
2a^2 - 14a - a + 7 = 0
\quad \Rightarrow \quad
(2a - 1)(a - 7) = 0.
Thus, a = \frac{1}{2} or a = 7. The question states a > 1, hence
a = 7.
Step 5: Find b using the definite integral
Because the point (a,b) lies on the curve, we compute
b = \int_0^a (2t^2 - 15t + 10)\,dt.
Substituting a = 7,
b = \int_0^7 (2t^2 - 15t + 10)\,dt.
Let us evaluate this integral:
\[
\int (2t^2 - 15t + 10)\,dt
= \frac{2}{3}t^3 - \frac{15}{2}t^2 + 10t + C.
\]
So,
\[
b = \left[\frac{2}{3}t^3 - \frac{15}{2}t^2 + 10t \right]_0^7.
\]
At t = 7 :
\[
\frac{2}{3}(7^3)
= \frac{2}{3} \cdot 343
= \frac{686}{3},
\quad
- \frac{15}{2}(7^2)
= -\frac{15}{2} \cdot 49
= -\frac{735}{2},
\quad
+ 10 \cdot 7
= 70.
\]
Hence,
\[
b = \left(\frac{686}{3}\right)
- \left(\frac{735}{2}\right)
+ 70.
\]
Converting to a common denominator of 6:
\[
\frac{686}{3} = \frac{1372}{6},
\quad
\frac{735}{2} = \frac{2205}{6},
\quad
70 = \frac{420}{6}.
\]
Thus,
\[
b = \frac{1372}{6} - \frac{2205}{6} + \frac{420}{6}
= \frac{1372 - 2205 + 420}{6}
= \frac{1372 + 420 - 2205}{6}
= \frac{1792 - 2205}{6}
= \frac{-413}{6}.
\]
Step 6: Calculate the required expression |a + 6b|
Since b = -\frac{413}{6}, we have
6b = -413.
Therefore,
a + 6b = 7 - 413 = -406.
Taking the absolute value,
|a + 6b| = |-406| = 406.
Answer: 406
