Question

Given \(\int_1^a (2x + 1) \, dx = 5\), find the sum of all values of \(a\).

Hint:

Start by applying the Fundamental Theorem of Calculus: integrate the function and substitute the limits so the given equation turns into a polynomial in the unknown limit. Once that polynomial is a quadratic, recall that there is a direct link between its coefficients and the sum of its roots, so each root does not need to be found separately.

Answer 1

Step 1: Understanding the Concept:
This is a definite integral problem where the result of the integration is given as a constant. We need to evaluate the integral in terms of the upper limit \(a\) and solve the resulting quadratic equation.

Step 2: Key Formula or Approach:
1. Power rule of integration: \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\).
2. Fundamental Theorem of Calculus: \(\int_p^q f(x) \, dx = [F(x)]_p^q = F(q) - F(p)\).
3. Sum of roots of a quadratic equation \(Ax^2 + Bx + C = 0\) is \(-B/A\).

Step 3: Detailed Explanation:
First, evaluate the integral:
\[ \int_1^a (2x + 1) \, dx = [x^2 + x]_1^a \] Applying the limits:
\[ (a^2 + a) - (1^2 + 1) = a^2 + a - 2 \] Given that the integral equals 5:
\[ a^2 + a - 2 = 5 \] \[ a^2 + a - 7 = 0 \] This is a quadratic equation in \(a\). We need to find the sum of all possible values of \(a\).
By comparing with \(Ax^2 + Bx + C = 0\), we have \(A = 1\) and \(B = 1\).
Sum of roots (values of \(a\)) \(= -\frac{B}{A} = -\frac{1}{1} = -1\).

Step 4: Final Answer:
The sum of all values of \(a\) is \(-1\).

Answer 2

Concept:

• A definite integral with an unknown limit evaluates to a specific number; substituting the limits into the antiderivative converts this into an ordinary algebraic equation in the unknown limit.
• Once that equation is a quadratic, both of its roots can be obtained directly from the quadratic formula, and then added together, instead of relying on a shortcut formula for the sum of roots.

Step 1: Evaluate the integral and form the equation.
$\int_1^a (2x+1)\,dx = [x^2+x]_1^a = (a^2+a) - (1+1) = a^2+a-2$
Setting this equal to the given value 5:
$a^2+a-2 = 5$
$a^2+a-7 = 0$

Step 2: Solve the quadratic equation using the quadratic formula.
For $Ax^2+Bx+C=0$, the roots are $a = \dfrac{-B \pm \sqrt{B^2-4AC}}{2A}$.
Here $A=1, B=1, C=-7$, so the discriminant is $D = 1^2-4(1)(-7) = 1+28 = 29$.
$a = \dfrac{-1 \pm \sqrt{29}}{2}$

Step 3: Add the two roots directly.
$a_1 = \dfrac{-1+\sqrt{29}}{2}$, $a_2 = \dfrac{-1-\sqrt{29}}{2}$
$a_1+a_2 = \dfrac{(-1+\sqrt{29})+(-1-\sqrt{29})}{2} = \dfrac{-2}{2} = -1$

Final Answer: The sum of all values of $a$ is $-1$.