Answer :
To evaluate
[tex]$$
\int_{-4}^2 \left(5x^6+7x^4\right)dx,
$$[/tex]
we will first find the antiderivative of the integrand and then use the Fundamental Theorem of Calculus.
1. Find the antiderivative:
For the term [tex]$5x^6$[/tex], we integrate as follows:
[tex]$$
\int 5x^6\,dx = 5 \cdot \frac{x^7}{7} = \frac{5x^7}{7}.
$$[/tex]
For the term [tex]$7x^4$[/tex], we get:
[tex]$$
\int 7x^4\,dx = 7 \cdot \frac{x^5}{5} = \frac{7x^5}{5}.
$$[/tex]
Thus, the antiderivative [tex]$F(x)$[/tex] is given by:
[tex]$$
F(x)=\frac{5x^7}{7}+\frac{7x^5}{5}.
$$[/tex]
2. Evaluate the antiderivative at the limits:
We now compute [tex]$F(2)$[/tex] and [tex]$F(-4)$[/tex].
- At the upper limit, [tex]$x=2$[/tex]:
[tex]$$
F(2)=\frac{5(2)^7}{7}+\frac{7(2)^5}{5}.
$$[/tex]
First, calculate the required powers:
- [tex]$(2)^7=128$[/tex], so [tex]$\frac{5\cdot128}{7}=\frac{640}{7}$[/tex].
- [tex]$(2)^5=32$[/tex], so [tex]$\frac{7\cdot32}{5}=\frac{224}{5}$[/tex].
To add these two fractions, express them with a common denominator, [tex]$35$[/tex]:
[tex]$$
\frac{640}{7}=\frac{640\times5}{35}=\frac{3200}{35},\quad \frac{224}{5}=\frac{224\times7}{35}=\frac{1568}{35}.
$$[/tex]
Hence,
[tex]$$
F(2)=\frac{3200}{35}+\frac{1568}{35}=\frac{4768}{35}.
$$[/tex]
- At the lower limit, [tex]$x=-4$[/tex]:
[tex]$$
F(-4)=\frac{5(-4)^7}{7}+\frac{7(-4)^5}{5}.
$$[/tex]
Compute the powers:
- [tex]$(-4)^7 = -16384$[/tex], so [tex]$\frac{5\cdot(-16384)}{7}=\frac{-81920}{7}$[/tex].
- [tex]$(-4)^5 = -1024$[/tex], so [tex]$\frac{7\cdot(-1024)}{5}=\frac{-7168}{5}$[/tex].
Convert these to have the same denominator, [tex]$35$[/tex]:
[tex]$$
\frac{-81920}{7}=\frac{-81920\times5}{35}=\frac{-409600}{35},\quad \frac{-7168}{5}=\frac{-7168\times7}{35}=\frac{-50176}{35}.
$$[/tex]
Therefore,
[tex]$$
F(-4)=\frac{-409600}{35}+\frac{-50176}{35}=\frac{-459776}{35}.
$$[/tex]
3. Compute the definite integral:
Using the Fundamental Theorem of Calculus,
[tex]$$
\int_{-4}^2\left(5x^6+7x^4\right)dx = F(2)-F(-4).
$$[/tex]
Substitute the values obtained:
[tex]$$
F(2)-F(-4)=\frac{4768}{35}-\left(\frac{-459776}{35}\right)=\frac{4768+459776}{35}=\frac{464544}{35}.
$$[/tex]
Thus, the value of the integral is:
[tex]$$
\boxed{\frac{464544}{35}}.
$$[/tex]
[tex]$$
\int_{-4}^2 \left(5x^6+7x^4\right)dx,
$$[/tex]
we will first find the antiderivative of the integrand and then use the Fundamental Theorem of Calculus.
1. Find the antiderivative:
For the term [tex]$5x^6$[/tex], we integrate as follows:
[tex]$$
\int 5x^6\,dx = 5 \cdot \frac{x^7}{7} = \frac{5x^7}{7}.
$$[/tex]
For the term [tex]$7x^4$[/tex], we get:
[tex]$$
\int 7x^4\,dx = 7 \cdot \frac{x^5}{5} = \frac{7x^5}{5}.
$$[/tex]
Thus, the antiderivative [tex]$F(x)$[/tex] is given by:
[tex]$$
F(x)=\frac{5x^7}{7}+\frac{7x^5}{5}.
$$[/tex]
2. Evaluate the antiderivative at the limits:
We now compute [tex]$F(2)$[/tex] and [tex]$F(-4)$[/tex].
- At the upper limit, [tex]$x=2$[/tex]:
[tex]$$
F(2)=\frac{5(2)^7}{7}+\frac{7(2)^5}{5}.
$$[/tex]
First, calculate the required powers:
- [tex]$(2)^7=128$[/tex], so [tex]$\frac{5\cdot128}{7}=\frac{640}{7}$[/tex].
- [tex]$(2)^5=32$[/tex], so [tex]$\frac{7\cdot32}{5}=\frac{224}{5}$[/tex].
To add these two fractions, express them with a common denominator, [tex]$35$[/tex]:
[tex]$$
\frac{640}{7}=\frac{640\times5}{35}=\frac{3200}{35},\quad \frac{224}{5}=\frac{224\times7}{35}=\frac{1568}{35}.
$$[/tex]
Hence,
[tex]$$
F(2)=\frac{3200}{35}+\frac{1568}{35}=\frac{4768}{35}.
$$[/tex]
- At the lower limit, [tex]$x=-4$[/tex]:
[tex]$$
F(-4)=\frac{5(-4)^7}{7}+\frac{7(-4)^5}{5}.
$$[/tex]
Compute the powers:
- [tex]$(-4)^7 = -16384$[/tex], so [tex]$\frac{5\cdot(-16384)}{7}=\frac{-81920}{7}$[/tex].
- [tex]$(-4)^5 = -1024$[/tex], so [tex]$\frac{7\cdot(-1024)}{5}=\frac{-7168}{5}$[/tex].
Convert these to have the same denominator, [tex]$35$[/tex]:
[tex]$$
\frac{-81920}{7}=\frac{-81920\times5}{35}=\frac{-409600}{35},\quad \frac{-7168}{5}=\frac{-7168\times7}{35}=\frac{-50176}{35}.
$$[/tex]
Therefore,
[tex]$$
F(-4)=\frac{-409600}{35}+\frac{-50176}{35}=\frac{-459776}{35}.
$$[/tex]
3. Compute the definite integral:
Using the Fundamental Theorem of Calculus,
[tex]$$
\int_{-4}^2\left(5x^6+7x^4\right)dx = F(2)-F(-4).
$$[/tex]
Substitute the values obtained:
[tex]$$
F(2)-F(-4)=\frac{4768}{35}-\left(\frac{-459776}{35}\right)=\frac{4768+459776}{35}=\frac{464544}{35}.
$$[/tex]
Thus, the value of the integral is:
[tex]$$
\boxed{\frac{464544}{35}}.
$$[/tex]
