5. Compute the given definite integrals. \begin{array}{ll} ext { (a) } \int_{-1}^{3}\left(3 x^{2}-2 x+1 ight) d x & ext { (d) } \int_{-1}^{1}\left(2-3 x+\frac{1}{x^{4}} ight) d x ext { (c) } \int_{3}^{8}\left(x^{3}+4 ight) d x & ext { Given that } \frac{d y}{d x}=2 x+7 ext { and } y=5 ext { when } x=-1 ext {, find the value of } y ext { when } x=2 . \end{array}

Question

5. Compute the given definite integrals. \begin{array}{ll}	ext { (a) } \int_{-1}^{3}\left(3 x^{2}-2 x+1ight) d x & 	ext { (d) } \int_{-1}^{1}\left(2-3 x+\frac{1}{x^{4}}ight) d x 	ext { (c) } \int_{3}^{8}\left(x^{3}+4ight) d x & 	ext { Given that } \frac{d y}{d x}=2 x+7 	ext { and } y=5 	ext { when } x=-1 	ext {, find the value of } y 	ext { when } x=2 . \end{array}

UpStudy AI · Accepted Answer

Let's tackle each part of the problem step by step.

---

## **5. Compute the given definite integrals.**

### **(a) $$\displaystyle \int_{-1}^{3} \left(3x^{2} - 2x + 1\right) \, dx$$**

**Step 1: Find the Antiderivative**

$$
\begin{align*}
\int 3x^{2} \, dx &= x^{3} \
\int -2x \, dx &= -x^{2} \
\int 1 \, dx &= x \
\end{align*}
$$

So, the antiderivative $$F(x)$$ is:

$$
F(x) = x^{3} - x^{2} + x
$$

**Step 2: Evaluate from $$-1$$ to $$3$$**

$$
\begin{align*}
F(3) &= 3^{3} - 3^{2} + 3 = 27 - 9 + 3 = 21 \
F(-1) &= (-1)^{3} - (-1)^{2} + (-1) = -1 - 1 - 1 = -3 \
\end{align*}
$$

**Step 3: Compute the Definite Integral**

$$
\int_{-1}^{3} \left(3x^{2} - 2x + 1\right) \, dx = F(3) - F(-1) = 21 - (-3) = 24
$$

---

### **(c) $$\displaystyle \int_{3}^{8} \left(x^{3} + 4\right) \, dx$$**

**Step 1: Find the Antiderivative**

$$
\begin{align*}
\int x^{3} \, dx &= \frac{1}{4}x^{4} \
\int 4 \, dx &= 4x \
\end{align*}
$$

So, the antiderivative $$F(x)$$ is:

$$
F(x) = \frac{1}{4}x^{4} + 4x
$$

**Step 2: Evaluate from $$3$$ to $$8$$**

$$
\begin{align*}
F(8) &= \frac{1}{4}(8)^{4} + 4(8) = \frac{1}{4}(4096) + 32 = 1024 + 32 = 1056 \
F(3) &= \frac{1}{4}(3)^{4} + 4(3) = \frac{1}{4}(81) + 12 = 20.25 + 12 = 32.25 \
\end{align*}
$$

**Step 3: Compute the Definite Integral**

$$
\int_{3}^{8} \left(x^{3} + 4\right) \, dx = F(8) - F(3) = 1056 - 32.25 = 1023.75 \quad \text{or} \quad \frac{4095}{4}
$$

---

### **(d) $$\displaystyle \int_{-1}^{1} \left(2 - 3x + \frac{1}{x^{4}}\right) \, dx$$**

**Step 1: Identify Potential Issues**

The integrand $$\frac{1}{x^{4}}$$ is undefined at $$x = 0$$. Therefore, the integral is **improper** and must be evaluated as a limit.

**Step 2: Split the Integral**

$$
\int_{-1}^{1} \left(2 - 3x + \frac{1}{x^{4}}\right) \, dx = \lim_{a \to 0^{-}} \int_{-1}^{a} \left(2 - 3x + \frac{1}{x^{4}}\right) \, dx + \lim_{b \to 0^{+}} \int_{b}^{1} \left(2 - 3x + \frac{1}{x^{4}}\right) \, dx
$$

**Step 3: Find the Antiderivative**

$$
\begin{align*}
\int 2 \, dx &= 2x \
\int -3x \, dx &= -\frac{3}{2}x^{2} \
\int \frac{1}{x^{4}} \, dx &= \int x^{-4} \, dx = -\frac{1}{3x^{3}} \
\end{align*}
$$

So, the antiderivative $$F(x)$$ is:

$$
F(x) = 2x - \frac{3}{2}x^{2} - \frac{1}{3x^{3}}
$$

**Step 4: Evaluate Each Part**

1. **As $$a \to 0^{-}$$:**

$$
   \lim_{a \to 0^{-}} \left[ F(a) - F(-1) \right] = \lim_{a \to 0^{-}} \left( 2a - \frac{3}{2}a^{2} - \frac{1}{3a^{3}} + 3.1667 \right )
   $$
   
   The term $$-\frac{1}{3a^{3}}$$ approaches $$+\infty$$ as $$a \to 0^{-}$$.

2. **As $$b \to 0^{+}$$:**

$$
   \lim_{b \to 0^{+}} \left[ F(1) - F(b) \right] = \lim_{b \to 0^{+}} \left( 0.1667 - 2b + \frac{3}{2}b^{2} + \frac{1}{3b^{3}} \right )
   $$
   
   The term $$\frac{1}{3b^{3}}$$ approaches $$+\infty$$ as $$b \to 0^{+}$$, making the entire expression approach $$-\infty$$.

**Step 5: Conclusion**

Since one part of the integral approaches $$+\infty$$ and the other approaches $$-\infty$$, the integral **does not converge**. Therefore:

$$
\int_{-1}^{1} \left(2 - 3x + \frac{1}{x^{4}}\right) \, dx \quad \text{is **divergent** and does not exist.}
$$

---

## **Additional Problem**

**Given:** $$\displaystyle \frac{dy}{dx} = 2x + 7$$ and $$y = 5$$ when $$x = -1$$.

**Find:** The value of $$y$$ when $$x = 2$$.

**Step 1: Find the General Solution**

$$
\frac{dy}{dx} = 2x + 7 \quad \Rightarrow \quad y(x) = \int (2x + 7) \, dx = x^{2} + 7x + C
$$

**Step 2: Determine the Constant $$C$$**

Use the initial condition $$y(-1) = 5$$:

$$
5 = (-1)^{2} + 7(-1) + C \quad \Rightarrow \quad 5 = 1 - 7 + C \quad \Rightarrow \quad C = 11
$$

**Step 3: Write the Particular Solution**

$$
y(x) = x^{2} + 7x + 11
$$

**Step 4: Compute $$y(2)$$**

$$
y(2) = (2)^{2} + 7(2) + 11 = 4 + 14 + 11 = 29
$$

**Answer:** When $$x = 2$$, $$y = 29$$.

---

## **Summary of Answers**

- **(a)** $$\displaystyle \int_{-1}^{3} \left(3x^{2} - 2x + 1\right) \, dx = 24$$
  
- **(c)** $$\displaystyle \int_{3}^{8} \left(x^{3} + 4\right) \, dx = \frac{4095}{4} $$ or $$1023.75$$
  
- **(d)** $$\displaystyle \int_{-1}^{1} \left(2 - 3x + \frac{1}{x^{4}}\right) \, dx$$ **does not converge**.
  
- **Additional Problem:** $$y = 29$$ when $$x = 2$$.
- **(a)** The integral from $$-1$$ to $$3$$ of $$3x^{2} - 2x + 1$$ is **24**. - **(c)** The integral from $$3$$ to $$8$$ of $$x^{3} + 4$$ is **1023.75**. - **(d)** The integral from $$-1$$ to $$1$$ of $$2 - 3x + \frac{1}{x^{4}}$$ **does not exist** because it diverges. - **Additional Problem:** When $$x = 2$$, $$y = 29$$.

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