Question

Evaluate the integrals . $$ \int_{0}^{1}\left(x^{2}-4 y\right) d x, \text { where } x=t+1, y=t^{3}+4 $$

Short answer

The integral evaluates to \( \frac{44}{3} \).

Step-by-step solution

Variable Substitution

We are given the integrals with the substitutions that are needed: \( x = t + 1 \) and \( y = t^3 + 4 \). Let's substitute these expressions into the integral.

Substitute x and y

Substitute \( x \) and \( y \) into the integral: \[ and substitute: \ x^2 = (t+1)^2 = t^2 + 2t + 1 \ y = t^3 + 4.\\\int_0^1 \left(x^2 - 4y\right) dx = \int_0^1 ((t^2 + 2t + 1) - 4(t^3 + 4)) dt\\]

Simplify the Expression

Simplify the expression inside the integral: \[(t^2 + 2t + 1) - 4(t^3 + 4) = t^2 + 2t + 1 - 4t^3 - 16\]Combine like terms: \[= -4t^3 + t^2 + 2t - 15\]

Adjust the Integral Bounds

Determine the bounds for \( t \). When \( x = 0 \) (the lower bound on \( x \)), \( t+1 = 0 \) gives \( t = -1 \). When \( x = 1 \) (the upper bound on \( x \)), \( t+1 = 1 \) gives \( t = 0 \). Update the integral limits from \( -1 \) to \( 0 \).

Integrate the Function

Calculate the integral of \(-4t^3 + t^2 + 2t - 15\) from \( t = -1 \) to \( t = 0 \):\[\int_{-1}^{0} (-4t^3 + t^2 + 2t - 15) \, dt \]This becomes:\[\left[ -t^4 + \frac{t^3}{3} + t^2 - 15t \right]_{-1}^{0}\]

Evaluate the Integral

Calculate the value of: \[\left[ -t^4 + \frac{t^3}{3} + t^2 - 15t \right]_{-1}^{0}\] Evaluating at \( t = 0 \) yields zero. For \( t = -1 \): \[-(-1)^4 + \frac{(-1)^3}{3} + (-1)^2 - 15(-1)\]Which results in:\[-1 - \frac{1}{3} + 1 + 15 = 15 - \frac{1}{3} = \frac{44}{3}\]

Final Step: Conclusion of Integration

Thus, after evaluating the integral, its value is \( \frac{44}{3} \).

Key concepts

Variable Substitution

In the world of calculus integration, variable substitution is like a secret weapon. It helps simplify complex integrals by changing the variables used in the expression. In our exercise, we swap out variables from the integral with given functions of a new variable, say \( t \).

• Why Substitution?: It can convert a complicated integral into a simpler one, making it easier to evaluate.
• How it Works: Substitute the given expressions \( x = t + 1 \) and \( y = t^3 + 4 \) directly into the integral expression.
• After substitution, the integral becomes a new expression dependent on \( t \) instead of the original variables \( x \) and \( y \).

Substituting makes the process less confusing and more straightforward, letting us focus on the mathematics rather than getting tangled in algebra.

Integral Bounds

Integral bounds play a crucial role in solving definite integrals. They tell us the interval over which you need to evaluate the function. These bounds are the start and end points of integration.

• Original Bounds: The exercise starts with bounds from \( x = 0 \) to \( x = 1 \).
• Transforming Bounds: With variable substitution, the bounds also need updating. Solve \( x = t + 1 \) for \( t \) with the old bound values. When \( x = 0 \), \( t = -1 \). When \( x = 1 \), \( t = 0 \).
• Now, the bounds are from \( t = -1 \) to \( t = 0 \), perfectly aligned with the substituted expression.

Accurate bounds ensure the evaluated integral represents the area under the curve over the correct interval.

Definite Integral

Definite integrals provide us with the net area under a curve bounded by the integral limits. This concept is pivotal in numerous real-world applications like calculating distance, area, and even probability.

• Expression After Substitution: Our new integral expression becomes \( \int_{-1}^{0} (-4t^3 + t^2 + 2t - 15) \, dt \).
• Solving the Integral: Integrate term by term, considering the power reduction of \( t^n \) to \( \frac{t^{n+1}}{n+1} \).
• The definite integral results in a single numerical value, achieved by evaluating the integrated function at the provided bounds.

The definite integral tells us not just about the curve but quantified information about its behavior over a specified domain.

Polynomial Integration

Polynomial integration involves integrating expressions made up of sums of powers of a variable. Each power is treated individually during integration.

• Terms to Handle: Our example includes terms like \(-4t^3, t^2, 2t, \) and \(-15 \).
• Integration Rule: For every term \( at^n \), the integral is \( \frac{a}{n+1}t^{n+1} \).
• Break down the integral into individual integrations of each term.
• Evaluate the polynomial at the integral bounds to get the final result.

Through polynomial integration, integrating even complex-looking polynomial expressions becomes manageable and systematic.