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Evaluate. $$\int_{1}^{3} \int_{0}^{x} 2 e^{x^{2}} d y d x$$

Short Answer

Expert verified
The value of the integral is \ e^{9} - e\.

Step by step solution

01

- Understand the Integral

The given integral is a double integral over the region defined by the limits. Rewrite the integral as: \[\int_{1}^{3} \bigg[\int_{0}^{x} 2 e^{x^{2}} d y \bigg] d x\]. The integrand is a function of x alone, so integration with respect to y is straightforward.
02

- Integrate with Respect to y

Integrate the inner integral with respect to y:\[\int_{0}^{x} 2 e^{x^{2}} d y = 2 e^{x^{2}} \bigg[y \bigg]_{0}^{x} = 2 e^{x^{2}} x = 2x e^{x^{2}} \]. Hence, the integral reduces to \[\int_{1}^{3} 2x e^{x^{2}} d x \].
03

- Integrate with Respect to x

Evaluate the integral \[\int_{1}^{3} 2x e^{x^{2}} d x \]. This can be solved by recognizing that the integrand is easily integrable by substitution. Let \ u = x^{2}, \ dv = 2x dx\, then\ du = 2x dx \ and so the integral becomes\ \int e^{u} du\. Integrate to get \ e^{u} \bigg]_{1}^{3} \ which is the same as\ e^{x^{2}} \bigg]_{1}^{3}.
04

- Compute the Result

Substitute the limits into the integrated form: \[\bigg[ e^{x^{2}} \bigg]_{1}^{3} = e^{3^{2}} - e^{1^{2}} = e^{9} - e^{1} = e^{9} - e \].So the value of the given integral is \ e^{9} - e\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Substitution
Integration by substitution is a powerful technique used to simplify integrals. It involves changing variables to make the integral easier to evaluate.
When faced with a complicated integrand, identify a part of it that can be substituted with a new variable. This is usually a function and its derivative. Here, let's see how this technique is applied:
For the integral \(\begin{aligned} \int_{1}^{3} 2x e^{x^{2}} dx \end{aligned}\), note that \(2x dx\) is the derivative of \(x^2\). This hints at using substitution. Let \(u = x^2\), which gives \(du = 2x dx\).
The integral now becomes \(\begin{aligned} \int e^u du \end{aligned}\), which is straightforward. After integrating, don't forget to change back to the original variable by substituting \(u = x^2\).
Evaluating Definite Integrals
A definite integral has both upper and lower limits. This means after finding the antiderivative, you evaluate it at the given limits and subtract.
Let's break it down using our transformed integral: \(\begin{aligned} \bigg[ e^u \bigg]_{x=1}^{x=3} \end{aligned}\)
After applying substitution and integrating, we get \(\begin{aligned} e^{u} \bigg]_{1}^{9} \end{aligned}\).
To evaluate it, substitute the limits \(1\) and \(9\) into \(u\), giving \(\begin{aligned} e^9 - e^1 \end{aligned}\).
This final subtraction step yields the value of the definite integral.
Exponential Function
An exponential function is a mathematical function of the form \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to \(2.718\).
Exponential functions grow rapidly and have unique properties that make them essential in calculus.
In the given problem, \(e^{x^2}\) dominates the integrand. Knowing how to integrate exponential functions, especially with substitution, simplifies solving these types of integrals.
The exponent in \(e^{x^2}\) can be managed by changing variables, as seen in our substitution step. These techniques allow handling integrals of exponential functions more efficiently.

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Most popular questions from this chapter

Find \(f_{x x}, f_{x y}, f_{y x},\) and \(f_{y y}\) (Remember, \(f_{y x}\) means to differentiate with respect to \(y\) and then with respect to \(x\).) $$f(x, y)=x \ln y$$

Evaluate. $$\int_{0}^{2} \int_{0}^{x} e^{x+y} d y d x$$

Temperature-humidity heat index. In the summer, humidity interacts with the outdoor temperature, making a person feel hotter because of reduced heat loss from the skin caused by higher humidity. The temperature-humidity index, \(T_{\mathrm{h}},\) is what the temperature would have to be with no humidity in order to give the same heat effect. One index often used is given by $$T_{\mathrm{h}}=1.98 T-1.09(1-H)(T-58)-56.9$$ where \(T\) is the air temperature, in degrees Fahrenheit, and H is the relative humidity, expressed as a decimal. Find the temperature-humidity index in each case. Round to the nearest tenth of a degree. $$T=90^{\circ} \mathrm{F} \text { and } \mathrm{H}=90 \%$$

Find the volume of the solid capped by the surface \(z=1-y-x^{2}\) over the region bounded on the \(x y\) -plane by \(y=1-x^{2}, y=0, x=0,\) and \(x=1,\) by evaluating the integral $$\int_{0}^{1} \int_{0}^{1-x^{2}}\left(1-y-x^{2}\right) d y d x$$.

Temperature-humidity heat index. In the summer, humidity interacts with the outdoor temperature, making a person feel hotter because of reduced heat loss from the skin caused by higher humidity. The temperature-humidity index, \(T_{\mathrm{h}},\) is what the temperature would have to be with no humidity in order to give the same heat effect. One index often used is given by $$T_{\mathrm{h}}=1.98 T-1.09(1-H)(T-58)-56.9$$ where \(T\) is the air temperature, in degrees Fahrenheit, and H is the relative humidity, expressed as a decimal. Find the temperature-humidity index in each case. Round to the nearest tenth of a degree. $$T=78^{\circ} \mathrm{F} \text { and } \mathrm{H}=100 \%$$

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