Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $$\int_{0}^{4} \frac{x}{x^{2}+1} d x$$

Short answer

Question: Evaluate the definite integral $$\int_{0}^{4} \frac{x}{x^{2}+1} d x$$. Answer: The definite integral equals $$\frac{1}{2} (\ln{17})$$.

Step-by-step solution

Choose the appropriate substitution

We will choose the substitution $$u = x^2 + 1$$, as it can eliminate the denominator and simplify the integrand.

Compute the differential

To find differential, differentiate $$u$$ with respect to $$x$$ and write $$d u$$ in terms of $$d x$$. $$\frac{d u}{d x} = 2x \Rightarrow d u = 2x d x$$

Change the integral limits according to the substitution

Now we will change the integral limits according to the substitution. Our new limits will be for $$u$$, not for $$x$$. Replace $$x$$ in the substitution equation with the original limits and find new limits for $$u$$. Lower limit: When $$x = 0$$, $$u = (0)^2 + 1 = 1$$ Upper limit: When $$x = 4$$, $$u = (4)^2 + 1 = 17$$ So, our new integral limits are $$1$$ and $$17$$.

Perform the substitution in the integral and simplify it

Now we will perform the substitution in the integral, use the expressions $$u = x^2 + 1$$ and $$d u = 2x d x$$ to replace $$x$$ and $$d x$$ in the integral. $$\int_{0}^{4} \frac{x}{x^{2}+1} d x = \int_{1}^{17} \frac{1}{2} \frac{d u}{u}$$

Evaluate the definite integral after simplifying

Now we will evaluate the simplified definite integral: $$\int_{1}^{17} \frac{1}{2} \frac{d u}{u} = \frac{1}{2} \int_{1}^{17} \frac{d u}{u} = \frac{1}{2} [\ln{u}]_{1}^{17} = \frac{1}{2} (\ln{17} - \ln{1})$$

State the final answer

The final answer is: $$\int_{0}^{4} \frac{x}{x^{2}+1} d x = \frac{1}{2} (\ln{17} - \ln{1}) = \frac{1}{2} (\ln{17})$$

Key concepts

U-Substitution Method

When solving definite integrals, the u-substitution method is a powerful technique that simplifies the process of integration. This method, also known as integration by substitution, effectively changes the variable of integration to make the integral easier to solve.

To apply the u-substitution method, we choose a substitution for the variable inside the integral that will simplify the integrand. This involves selecting a part of the integrand to be our new variable u, and then expressing the differential dx in terms of du. It's essential to choose a substitution that makes the integral more manageable.

For example, in the given exercise, the substitution \( u = x^2 + 1 \) was chosen because it allows for the \( x \) in the numerator to be replaced and the differential \( dx \) can be rewritten in terms of \( du \). This is done by differentiating u with respect to x to get \( du = 2x dx \), which simplifies the integral drastically.

By replacing \( x \) and \( dx \) with their u-equivalents, the integrand becomes a simple function of u that can be easily integrated.

Change of Variables in Integrals

Changing variables in integrals is closely related to the u-substitution method and essentially serves the same purpose - to simplify the integral by altering its variable. When we perform a change of variables, it's crucial to also change the limits of integration to match the new variable.

In the provided exercise, after the substitution \( u = x^2 + 1 \) is made, the original integration limits which are in terms of x must be converted to new limits in terms of u. This is done by substituting the original x-values into the u-substitution equation.

For the lower limit when \( x = 0 \), the new limit becomes \( u = 1 \). Similarly, for the upper limit when \( x = 4 \), the new limit is \( u = 17 \). These new limits ensure that our integration bounds accurately reflect the interval over which we're integrating with respect to u.

Integral Evaluation

Once we have made the appropriate substitutions and changed the variables, evaluating the integral becomes the final step. This involves applying the fundamental theorem of calculus, which ties together the integral of a function with its antiderivative.

After the substitutions in the given exercise, the integral simplifies to \( \frac{1}{2} \int_{1}^{17} \frac{du}{u} \). The antiderivative of \( \frac{1}{u} \) is \( \text{ln}|u| \), where ln denotes the natural logarithm. To evaluate the definite integral, we find the antiderivative at both the upper and the lower limits and then subtract the two values.

The integral from 1 to 17 of \( \frac{1}{u} \) simplifies to \( \text{ln}(17) - \text{ln}(1) \), which further simplifies to \( \text{ln}(17) \) since the natural logarithm of 1 is 0. Multiplying by the pre-factor \( \frac{1}{2} \), we obtain the final value of the integral, which concludes the problem-solving process.