Question

Completing the square Evaluate the following integrals. $$\int_{1}^{4} \frac{d t}{t^{2}-2 t+10}$$

Short answer

Question: Evaluate the given integral: $$\int_{1}^{4} \frac{d t}{t^{2}-2 t+10}$$ Answer: $$\frac{\pi}{6}$$

Step-by-step solution

Complete the square

First, we need to complete the square for the quadratic expression in the denominator. The expression is \(t^2 - 2t + 10\). To complete the square, take half of the coefficient of the linear term (-2), square it, and add and subtract that value to the expression. $$t^2 - 2t + 10 = (t^2 - 2t + 1) + 9 = (t - 1)^2 + 9$$

Rewrite the integral

Now, we substitute the completed square form of the denominator back into the integral: $$\int_{1}^{4} \frac{d t}{(t - 1)^2 + 9}$$

Apply the trigonometric substitution

We can perform the trigonometric substitution by setting: $$t - 1 = 3\sin(\theta) \Rightarrow dt = 3\cos(\theta)d\theta$$ Now, replace \(t\) and \(dt\) with the trigonometric substitution: $$\int_{\theta_1}^{\theta_4} \frac{3\cos(\theta)}{(3\sin(\theta))^2 + 9}d\theta$$ Where \(\theta_1 = \sin^{-1}\left(\frac{1-1}{3}\right) = 0\) and \(\theta_4 = \sin^{-1}\left(\frac{4-1}{3}\right)\).

Simplify the integral expression

Simplify the obtained integral expression: $$\int_{0}^{\sin^{-1}(1)} \frac{3\cos(\theta)}{9\sin^2(\theta) + 9}d\theta = \int_{0}^{\sin^{-1}(1)} \frac{\cos(\theta)}{3\sin^2(\theta) + 3}d\theta$$ $$=\frac{1}{3}\int_{0}^{\sin^{-1}(1)} \frac{\cos(\theta)}{\sin^2(\theta) + 1}d\theta = \frac{1}{3}\int_{0}^{\sin^{-1}(1)} d\theta$$ By substituting back \(\theta\) to \(t\) (Inverse trig substitution): $$\frac{1}{3}\int_{0}^{\sin^{-1}(1)} d\theta = \frac{1}{3}[\theta]_{0}^{\sin^{-1}(1)} = \frac{1}{3}\left[\sin^{-1}(1)-0\right] = \frac{1}{3}\left[\frac{\pi}{2}\right]$$

Evaluate the integral

Finally, evaluate the integral: $$\frac{1}{3}\left[\frac{\pi}{2}\right] = \frac{\pi}{6}$$ Thus, the value of the integral is: $$\int_{1}^{4} \frac{d t}{t^{2}-2 t+10} = \frac{\pi}{6}$$

Key concepts

Completing the Square

Completing the square is a powerful algebraic technique used to transform a quadratic expression into a perfect square trinomial plus or minus a constant. This method not only aids in solving quadratic equations but also plays a significant role in calculus, especially in integration problems involving quadratic expressions.

To complete the square for a quadratic expression like \(t^2 - 2t + 10\), here’s what you do:

• Identify the coefficient of the linear term (here it's \(-2\)).
• Take half of that coefficient, giving you \(-1\).
• Square that result, resulting in \(1\).
• Add and subtract this square inside the expression.

Applying these steps to our expression, we have:
\[t^2 - 2t + 10 = (t - 1)^2 + 9\]
By converting the expression into this form, the integral becomes easier to handle, often allowing for trigonometric substitution.

Trigonometric Substitution

Trigonometric substitution is a method used in calculus to simplify integrals, especially those involving square roots or quadratic expressions. It leverages trigonometric identities for easier evaluation by changing variables.

For our integral:
- After completing the square, the denominator becomes \((t - 1)^2 + 9\), resembling the identity \(a^2 + x^2\).
- We use the substitution \(t - 1 = 3 \sin(\theta)\), so that the expression matches the identity. Here, \(3 heta\) aligns with \(a\cdot\sin(\theta)\).
- Differentiating, \(dt = 3\cos(\theta)d\theta\).

Substituting these into the integral, we can simplify:\[\int \frac{3 \cos(\theta)}{(3 \sin(\theta))^2 + 9}d\theta\]
With trigonometric identities, \(\sin^2(\theta) + \cos^2(\theta) = 1\), it simplifies to an easily integrable form.

Definite Integrals

Definite integrals are a staple in calculus, representing the signed area under a curve between two points. Unlike indefinite integrals, which have a constant of integration, definite integrals result in a real number.

They follow the form:
\[\int_a^b f(x)dx\]
Where \(a\) and \(b\) are the limits of integration.

In our exercise, the integral \(\int_{1}^{4} \frac{dt}{t^{2} - 2t + 10}\) has bounds from 1 to 4. Using our trigonometric substitution and simplifying the expression, we're then able to evaluate the integral from \(\theta = 0\) to \(\theta = \sin^{-1}(1)\), translating into exact limits in the \(\theta\) domain.

After calculating, we obtain the evaluated definite integral as \(\frac{\pi}{6}\), representing the exact area under the curve from \(t=1\) to \(t=4\). This process showcases the beauty and utility of definite integrals in finding precise areas under complex functions.