In Problems 61-68, find each definite integral two ways: (a) By finding the related indefinite integral and then using the Fundamental Theorem of Calculus; $$ \begin{array}{ll} ext { (b) By making a substitution in the integrand and using the } \ ext { substitution to change the limits of integration. } \ ext { (c) Which method did you prefer? Why? } \ ext { (438. 61. } \int_{-2}^{0} \frac{x}{\left(x^{2}+3 ight)^{2}} d x & ext { 62. } \int_{-1}^{1}\left(x^{2}-1 ight)^{5} x d x \ ext { 63. } \int_{0}^{1} x^{2} e^{x^{3}+1} d x & ext { 64. } \int_{0}^{1} x e^{x^{2}-2} d x \ ext { 65. } \int_{1}^{6} x \sqrt{x+3} d x & ext { 66. } \int_{2}^{6} x^{2} \sqrt{x-2} d x \ ext { 67. } \int_{0}^{2} x \cdot 3^{2 x^{2}} d x & ext { 68. } \int_{0}^{1} x \cdot 10^{-x^{2}} d x\end{array} $$

Question

In Problems 61-68, find each definite integral two ways: (a) By finding the related indefinite integral and then using the Fundamental Theorem of Calculus; $$ \begin{array}{ll}	ext { (b) By making a substitution in the integrand and using the } \ 	ext { substitution to change the limits of integration. } \ 	ext { (c) Which method did you prefer? Why? } \ 	ext { (438. 61. } \int_{-2}^{0} \frac{x}{\left(x^{2}+3ight)^{2}} d x & 	ext { 62. } \int_{-1}^{1}\left(x^{2}-1ight)^{5} x d x \ 	ext { 63. } \int_{0}^{1} x^{2} e^{x^{3}+1} d x & 	ext { 64. } \int_{0}^{1} x e^{x^{2}-2} d x \ 	ext { 65. } \int_{1}^{6} x \sqrt{x+3} d x & 	ext { 66. } \int_{2}^{6} x^{2} \sqrt{x-2} d x \ 	ext { 67. } \int_{0}^{2} x \cdot 3^{2 x^{2}} d x & 	ext { 68. } \int_{0}^{1} x \cdot 10^{-x^{2}} d x\end{array} $$

UpStudy AI · Accepted Answer

Let's solve the first integral from the list, which is:

$$
\int_{-2}^{0} \frac{x}{(x^{2}+3)^{2}} \, dx
$$

### Part (a): Finding the Indefinite Integral

To find the indefinite integral, we can use integration by substitution. Let:

$$
u = x^2 + 3 \implies du = 2x \, dx \implies dx = \frac{du}{2x}
$$

Now, we need to express $$x$$ in terms of $$u$$:

$$
x = \sqrt{u - 3} \quad \text{(we will consider the negative root since we are integrating from -2 to 0)}
$$

Now, substituting into the integral:

$$
\int \frac{x}{(x^2 + 3)^2} \, dx = \int \frac{x}{u^2} \cdot \frac{du}{2x} = \frac{1}{2} \int u^{-2} \, du
$$

Now, we can integrate:

$$
\frac{1}{2} \int u^{-2} \, du = \frac{1}{2} \left(-\frac{1}{u}\right) + C = -\frac{1}{2u} + C
$$

Substituting back for $$u$$:

$$
-\frac{1}{2(x^2 + 3)} + C
$$

Now, we can apply the Fundamental Theorem of Calculus:

$$
\int_{-2}^{0} \frac{x}{(x^2 + 3)^2} \, dx = \left[-\frac{1}{2(x^2 + 3)}\right]_{-2}^{0}
$$

Calculating the limits:

1. At $$x = 0$$:

$$
-\frac{1}{2(0^2 + 3)} = -\frac{1}{6}
$$

2. At $$x = -2$$:

$$
-\frac{1}{2((-2)^2 + 3)} = -\frac{1}{2(4 + 3)} = -\frac{1}{14}
$$

Now, substituting these values into the definite integral:

$$
\int_{-2}^{0} \frac{x}{(x^2 + 3)^2} \, dx = -\frac{1}{6} - \left(-\frac{1}{14}\right) = -\frac{1}{6} + \frac{1}{14}
$$

Finding a common denominator (42):

$$
-\frac{7}{42} + \frac{3}{42} = -\frac{4}{42} = -\frac{2}{21}
$$

### Part (b): Substitution with Changed Limits

Using the same substitution $$u = x^2 + 3$$, we change the limits of integration:

- When $$x = -2$$:

$$
u = (-2)^2 + 3 = 4 + 3 = 7
$$

- When $$x = 0$$:

$$
u = 0^2 + 3 = 3
$$

Now, the integral becomes:

$$
\int_{7}^{3} \frac{x}{u^2} \cdot \frac{du}{2x} = \frac{1}{2} \int_{7}^{3} u^{-2} \, du
$$

Reversing the limits gives us:

$$
-\frac{1}{2} \int_{3}^{7} u^{-2} \, du
$$

Now, integrating:

$$
-\frac{1}{2} \left[-\frac{1}{u}\right]_{3}^{7} = -\frac{1}{2} \left(-\frac{1}{7} + \frac{1}{3}\right) = \frac{1}{2} \left(\frac{1}{3} - \frac{1}{7}\right)
$$

Finding a common denominator (21):

$$
\frac{1}{2} \left(\frac{7}{21} - \frac{3}{21}\right) = \frac{1}{2} \cdot \frac{4}{21} = \frac{2}{21}
$$

### Conclusion

Both methods yield the same result:

$$
\int_{-2}^{0} \frac{x}{(x^2 + 3)^2} \, dx = -\frac{2}{21}
$$

### Part (c): Preference

I prefer the substitution method for this problem because it allows for a more straightforward calculation of the limits and avoids the need for additional algebraic manipulation. It also provides a clear view of how the function behaves over the interval.
To solve the integral $$ \int_{-2}^{0} \frac{x}{(x^{2}+3)^{2}} \, dx $$, we can use two methods: 1. **Indefinite Integral Approach**: - Let $$ u = x^2 + 3 $$, then $$ du = 2x \, dx $$. - The integral becomes $$ \frac{1}{2} \int u^{-2} \, du = -\frac{1}{2u} + C $$. - Substituting back, we get $$ -\frac{1}{2(x^2 + 3)} + C $$. - Applying the Fundamental Theorem of Calculus: $$ \left[-\frac{1}{2(x^2 + 3)}\right]_{-2}^{0} = -\frac{1}{6} - \left(-\frac{1}{14}\right) = -\frac{2}{21} $$ 2. **Substitution with Changed Limits**: - Using the same substitution $$ u = x^2 + 3 $$, the limits change from $$ x = -2 $$ to $$ u = 7 $$ and $$ x = 0 $$ to $$ u = 3 $$. - The integral becomes $$ \frac{1}{2} \int_{7}^{3} u^{-2} \, du = \frac{1}{2} \left[-\frac{1}{u}\right]_{7}^{3} = \frac{2}{21} $$. Both methods yield the same result: $$ -\frac{2}{21} $$. **Preference**: The substitution method is preferred for its simplicity and clarity in handling the limits of integration.

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