Mixing
Find the interval in which $int_{-1}^{x} (e^{t} -1)(2-t) dt , (x>-1)$.
$begingroup$
Find the interval in which $F(x) = int_{-1}^{x} (e^{t}-1)(2-t) dt$ , $(x>-1)$ is increasing.
By solving integral
$$int_{-1}^{x} ((2-t)e^{t} -2 +t)dt
= [(2-t)e^{t} + e^{t} -2t + t^{2}/2]_{-1}^{x}$$
and this gives me
$$3e^{x} -xe^{x} -2x + x^{2}/2 - 4/e - 5/2$$
now, i don't know how to proceed.
I don't how to find roots so that i can get the interval.
Is there some easy way to solve the problem?
calculus
edited Jan 1 at 11:49
Robert Z
94.2k1061132
asked Jan 1 at 11:39
MathsaddictMathsaddict
2908
$endgroup$
add a comment |
$begingroup$
Find the interval in which $F(x) = int_{-1}^{x} (e^{t}-1)(2-t) dt$ , $(x>-1)$ is increasing.
By solving integral
$$int_{-1}^{x} ((2-t)e^{t} -2 +t)dt
= [(2-t)e^{t} + e^{t} -2t + t^{2}/2]_{-1}^{x}$$
and this gives me
$$3e^{x} -xe^{x} -2x + x^{2}/2 - 4/e - 5/2$$
now, i don't know how to proceed.
I don't how to find roots so that i can get the interval.
Is there some easy way to solve the problem?
calculus
edited Jan 1 at 11:49
Robert Z
94.2k1061132
asked Jan 1 at 11:39
MathsaddictMathsaddict
2908
$endgroup$
add a comment |
$begingroup$
Find the interval in which $F(x) = int_{-1}^{x} (e^{t}-1)(2-t) dt$ , $(x>-1)$ is increasing.
By solving integral
$$int_{-1}^{x} ((2-t)e^{t} -2 +t)dt
= [(2-t)e^{t} + e^{t} -2t + t^{2}/2]_{-1}^{x}$$
and this gives me
$$3e^{x} -xe^{x} -2x + x^{2}/2 - 4/e - 5/2$$
now, i don't know how to proceed.
I don't how to find roots so that i can get the interval.
Is there some easy way to solve the problem?
calculus
edited Jan 1 at 11:49
Robert Z
94.2k1061132
asked Jan 1 at 11:39
MathsaddictMathsaddict
2908
$endgroup$
Find the interval in which $F(x) = int_{-1}^{x} (e^{t}-1)(2-t) dt$ , $(x>-1)$ is increasing.
By solving integral
$$int_{-1}^{x} ((2-t)e^{t} -2 +t)dt
= [(2-t)e^{t} + e^{t} -2t + t^{2}/2]_{-1}^{x}$$
and this gives me
$$3e^{x} -xe^{x} -2x + x^{2}/2 - 4/e - 5/2$$
now, i don't know how to proceed.
I don't how to find roots so that i can get the interval.
Is there some easy way to solve the problem?
calculus
calculus
edited Jan 1 at 11:49
Robert Z
94.2k1061132
asked Jan 1 at 11:39
MathsaddictMathsaddict
2908
edited Jan 1 at 11:49
Robert Z
94.2k1061132
asked Jan 1 at 11:39
MathsaddictMathsaddict
2908
edited Jan 1 at 11:49
Robert Z
94.2k1061132
edited Jan 1 at 11:49
Robert Z
94.2k1061132
edited Jan 1 at 11:49
Robert Z
94.2k1061132
94.2k1061132
asked Jan 1 at 11:39
MathsaddictMathsaddict
2908
asked Jan 1 at 11:39
MathsaddictMathsaddict
2908
asked Jan 1 at 11:39
MathsaddictMathsaddict
2908
2908
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
$$f(x)=(e^x-1)(2-x),$$
so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.
answered Jan 1 at 11:45
Robert ZRobert Z
94.2k1061132
$endgroup$
1
$begingroup$
Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
$endgroup$
– Mathsaddict
Jan 1 at 11:51
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
$$f(x)=(e^x-1)(2-x),$$
so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.
answered Jan 1 at 11:45
Robert ZRobert Z
94.2k1061132
$endgroup$
1
$begingroup$
Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
$endgroup$
– Mathsaddict
Jan 1 at 11:51
add a comment |
$begingroup$
Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
$$f(x)=(e^x-1)(2-x),$$
so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.
answered Jan 1 at 11:45
Robert ZRobert Z
94.2k1061132
$endgroup$
1
$begingroup$
Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
$endgroup$
– Mathsaddict
Jan 1 at 11:51
add a comment |
$begingroup$
Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
$$f(x)=(e^x-1)(2-x),$$
so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.
answered Jan 1 at 11:45
Robert ZRobert Z
94.2k1061132
$endgroup$
Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
$$f(x)=(e^x-1)(2-x),$$
so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.
answered Jan 1 at 11:45
Robert ZRobert Z
94.2k1061132
answered Jan 1 at 11:45
Robert ZRobert Z
94.2k1061132
answered Jan 1 at 11:45
Robert ZRobert Z
94.2k1061132
answered Jan 1 at 11:45
Robert ZRobert Z
94.2k1061132
94.2k1061132
1
$begingroup$
Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
$endgroup$
– Mathsaddict
Jan 1 at 11:51
add a comment |
1
$begingroup$
Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
$endgroup$
– Mathsaddict
Jan 1 at 11:51
1
1
$begingroup$
Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
$endgroup$
– Mathsaddict
Jan 1 at 11:51
$begingroup$
Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
$endgroup$
– Mathsaddict
Jan 1 at 11:51
add a comment |
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