Problem 9 - Calculus Single Variable

Question

Suppose that \(\int_{5}^{-3}(-3 f(x) / 4) d x=7\). Evaluate \(\int_{5}^{-3}(6 f(x)+1) d x\).

Step-by-Step Solution

Answer

The value of the integral is -64.

Step 1: Understand the given integral

You're given the integral \( \int_{5}^{-3}\left(-\frac{3}{4} f(x)\right) dx = 7 \). This means the integral of the function \(-\frac{3}{4}f(x)\) from 5 to -3 is equal to 7.

Step 2: Break down the target integral

The integral we need to evaluate is \( \int_{5}^{-3}(6 f(x) + 1) dx \). This can be broken down into two separate integrals: \( \int_{5}^{-3}6 f(x) dx \) and \( \int_{5}^{-3}1 dx \).

Step 3: Solve for \(\int_{5}^{-3} f(x) dx\)

Given \( \int_{5}^{-3}\left(-\frac{3}{4} f(x)\right) dx = 7 \), we can rearrange to find \( \int_{5}^{-3} f(x) dx \). Let \( I = \int_{5}^{-3} f(x) dx \). Then \( -\frac{3}{4}I = 7 \). Solving for \( I \), we get \( I = -\frac{7 \times 4}{3} = -\frac{28}{3} \).

Step 4: Scale \(\int_{5}^{-3} f(x) dx\) by 6

We know \( \int_{5}^{-3} f(x) dx = -\frac{28}{3} \). Now calculate \( \int_{5}^{-3} 6 f(x) dx = 6 \int_{5}^{-3} f(x) dx \). Thus, \( \int_{5}^{-3} 6 f(x) dx = 6 \times -\frac{28}{3} = -56 \).

Step 5: Calculate \(\int_{5}^{-3} 1 dx\)

Since the integral of 1 is just the difference of the bounds, \( \int_{5}^{-3} 1 dx = (-3) - 5 = -8 \).

Step 6: Add the results

Combine the results from steps 4 and 5: \( \int_{5}^{-3}(6 f(x) + 1)dx = \int_{5}^{-3}6 f(x) dx + \int_{5}^{-3}1 dx = -56 - 8 = -64 \).

Key Concepts

Definite Integral

Understanding definite integrals is essential in integral calculus. A definite integral is used to find the accumulated value of a function, such as area under a curve, between two bounds.
When evaluating a definite integral, the limits of integration are crucial. These limits tell you from where to where the accumulation (like area) should be calculated.
For example, in the original exercise, we have \[ \int_{5}^{-3} \left(-\frac{3}{4} f(x)\right) dx = 7 \] Here, the bounds are from 5 to -3, and the function's integral gives you 7.

• This signifies that the effect of the function, when adjusted by the factor \(-\frac{3}{4}\), leads to a net result of 7 over the interval.
• Notice that flipping the bounds (from 5 to -3 instead of -3 to 5) changes the sign of the definite integral.

The operation of the definite integral usually results in a numerical value rather than a new function, highlighting how much a quantity has changed or been accumulated between two points of interest.

Integration Techniques

Integration techniques are the mathematical methods used to solve integrals.
In this exercise, breaking down the function is a primary technique. We see it with
\[ \int_{5}^{-3} (6 f(x) + 1) dx \] This expression is divided into two separate integrals:

• \( \int_{5}^{-3} 6 f(x) dx \)
• \( \int_{5}^{-3} 1 dx \)

With this approach, we deal with simpler problems that can be handled more easily.Additionally, scaling the integral is another technique used. We solve for \( \int_{5}^{-3} f(x) dx \), given a multiplier, using the provided integral:

• \( -\frac{3}{4} \int_{5}^{-3} f(x) dx = 7 \)
• Solve for the original: \( \int_{5}^{-3} f(x) dx = -\frac{28}{3} \)

Then, by scaling with 6, we find \( \int_{5}^{-3} 6 f(x) dx = -56 \). Breaking down integrals into parts and scaling them are key techniques as they simplify complex integrals, making them approachable.

Calculus Problem Solving

Problem solving in calculus involves strategic thinking and the application of various integration methods. It requires a step-by-step approach to deconstruct complex problems.In our solved exercise, unraveling a composite integral scenario was crucial.

• Firstly, identify what is given: the integral involving a scaled version of the unknown function.
• Next, find integrated values by rearranging equations; factor in constants and scalars wisely.

For example, using the equation \( -\frac{3}{4}\int_{5}^{-3} f(x) dx = 7 \), we isolate \( \int_{5}^{-3} f(x) dx \), enabling further calculations.Moreover, understanding the simple operation of the bounds, as in \( \int_{5}^{-3} 1 dx = -8 \), can aid in finding precise solutions when parts are broken down.
Combining results is the final step that provides the full picture:
\( \int_{5}^{-3} (6f(x) + 1) dx = -64 \). Effective calculus problem solving uses a clear breakdown of tasks, allowing for integration into streamlined, logical solutions. Applying theory to practice ensures clarity and correctness in results.