91Ó°ÊÓ

The definite integral allows us to calculate the area under a curve within specific bounds, offering a snapshot of a function over a given interval. When we evaluate the definite integral of a function, we look at the accumulation of quantities, which could represent physical concepts like distance. For example, solving \[ \int_{a}^{b} f(x) \, dx \] determines the net change of the function between the limits \( a \) and \( b \).With definite integrals:

• The lower limit of integration (often denoted as \( a \)) indicates where calculation starts.
• The upper limit (denoted as \( b \)) is where it ends.

The process results in numerical values rather than functions, giving insight into the size of the area or quantity represented by \( f(x) \) over the interval \([a, b]\).
In our exercise, the definite integral was from \(-2\)to \(2\), resulting in the calculated area of \(\frac{\pi}{4}\).

The Fundamental Theorem of Calculus bridges two main concepts: differentiation and integration. This theorem is significant because it allows us to compute definite integrals using antiderivatives, which simplifies complex calculations.This theorem consists of two parts:

• The first part informs us that if \( F \) is an antiderivative of \( f \), then the definite integral of \( f \) from \( a \)to \( b \)can be expressed as \( F(b) - F(a) \).
• The second part involves differentiating the integral of a function to return to the original function.

In our context, after finding the antiderivative \( \frac{1}{2}\arctan\left(\frac{x}{2}\right) \),we applied this theorem to quickly compute the definite integral by evaluating the antiderivative at the upper limit, \( 2, \)and subtracting its evaluation at the lower limit, \(-2\).
This displayed the power of the theorem by providing the quick solution \( \frac{\pi}{4} \).

Inverse trigonometric functions are essential when dealing with integrals involving expressions like \( \frac{1}{a^2+x^2} \).These functions allow us to find the angle whose trigonometric value is a given number.The integral \( \int \frac{1}{a^2 + x^2} \, dx \)ighlights the role of the arctangent or inverse tangent function, \( \arctan(x) \).The derivative of \( \arctan(x) \)is \( \frac{1}{1+x^2} \),letting us recognize patterns in integrals that involve sums of squares.To solve such integrals:

• Identify the constant \( a \) so that your integrand matches the form \( \frac{1}{a^2 + x^2} \).
• Recall the inverse trigonometric function that pairs with this form, which is \( \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C \).

Applying this to \( \int \frac{1}{4 + x^2} \, dx \)showed that \( a = 2 \),leading us to compute \( \frac{1}{2}\arctan\left(\frac{x}{2}\right) + C \),which in turn, facilitated evaluating the definite integral.