Question

Evaluate the following integrals. $$\int_{0}^{\sqrt{3}} \frac{6 x^{3}}{\sqrt{x^{2}+1}} d x$$

Short answer

Answer: The value of the integral is 3.

Step-by-step solution

Introduce the integral notation

We are given the following integral to evaluate: $$\int_{0}^{\sqrt{3}} \frac{6 x^{3}}{\sqrt{x^{2}+1}} d x$$

Perform substitution

To simplify the expression in the integral, let's use the substitution method. We set: $$u = x^2 + 1$$ Then, differentiate with respect to \(x\): $$\frac{d u}{d x} = 2x$$ Now, we can solve for \(dx\): $$d x = \frac{d u}{2x}$$ Because we have a definite integral, we also need to change the limits of integration. From the substitution \(u = x^2 + 1\), we get: - When \(x = 0, u = 0^2 + 1 = 1\) - When \(x = \sqrt{3}, u = (\sqrt{3})^2 + 1 = 3 + 1 = 4\) Our integral becomes: $$\int_{1}^{4} \frac{6 x^{3}}{\sqrt{u}} \left(\frac{d u}{2x}\right)$$

Solve the integral

Simplify the expression in the integral and cancel out common factors: $$\int_{1}^{4} \frac{3x^{2}}{\sqrt{u}} d u$$ Now the integral is easier to solve: $$\int_{1}^{4} \frac{3}{u^{\frac{1}{2}}} d u$$ $$= 3 \int_{1}^{4} u^{-\frac{1}{2}} d u$$ Now we use the power rule, which states that 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 of \(u^n\) with respect to \(u\) is \(u^{n+1}/(n+1) + C\): $$= 3u^{\frac{1}{2}} \Big|_1^4$$

Apply the fundamental theorem of calculus

Now, we apply the fundamental theorem of calculus by subtracting the original lower limit of integration from the upper limit of integration: $$= 3\left[4^{\frac{1}{2}} - 1^{\frac{1}{2}}\right]$$

Compute the final expression

Finally, we compute the final expression: $$= 3\left[2 - 1\right] = 3 \cdot 1 = 3$$ So, the value of the integral is: $$\int_{0}^{\sqrt{3}} \frac{6 x^{3}}{\sqrt{x^{2}+1}} d x = 3$$

Key concepts

Integration by Substitution

The technique of integration by substitution is similar to the 'u-substitution' method used in differentiation. It is particularly useful when dealing with functions that are products of a function and its derivative, or where a substitution can simplify the integral into a more recognizable form.

To execute integration by substitution, you first identify a portion of the integral that can be set as a new variable, usually denoted as u. Next, you differentiate this new variable with respect to the original variable to find du. After the substitution, you end up with an integral in terms of u, which is often easier to integrate. As seen in the exercise, once the substitution is done, and the integral is resolved, the final integral looks simpler and more familiar, easing the process of integration.

Fundamental Theorem of Calculus

A pivotal concept in integral calculus is the Fundamental Theorem of Calculus, which bridges the concept of differentiation and integration. It asserts that if a function f(x) is continuous on an interval [a, b], and F is an antiderivative of f on [a, b], then the definite integral of f from a to b is equal to F(b) - F(a).

Applying this theorem allows us to evaluate definite integrals by finding the antiderivative of the function. You used this theorem in the exercise by first determining the antiderivative and then computing the difference F(b) - F(a), which gave you the exact area under the curve from a to b.

Power Rule for Integration

The power rule for integration is a crucial tool when dealing with polynomials. It states that for any real number n that is not equal to -1, the antiderivative of x^n with respect to x is x^(n+1)/(n+1) plus a constant of integration C. In your solution, you employed the power rule to integrate u^(-1/2) which became u^(1/2)/(1/2) + C or, after simplifying, 2u^(1/2) + C. This rule greatly simplified the process and led you directly to find the antiderivative of the integrand.

Indefinite Integrals

An indefinite integral represents the antiderivative of a function. It includes a constant of integration C because the derivative of a constant is zero and hence the original function could have been any function with that derivative plus a constant.

While definite integrals give a numerical result representing the area under a curve, indefinite integrals result in a function plus a constant. In the step by step solution provided in the exercise, you were actually finding an indefinite integral after the substitution, before applying the limits of the definite integral to find the specific area under the curve.