It is the derivative of the function at the given value
y = x2 + 4x at x = −5
y = x2 + 4x at x = −5
dx/dy x=-5 =-6
What is -6
It is the Anti Derivative of:
3x2
What is x3 + c
It is the evaluation of the definite integral.
y = − x2/2 + x − 1/2 ; [−2, 1]
What is -1/2
It is the area under the curve over the given interval.
y=4/x2 ; [-2,-1]
I-1-2 4/x2 dx
What is 2
It is the U value of the substitution.
I20 2x[x2+4]2dx
What is u=x2+4
It is the slope of the function at the given value.
y = x2 + 6x + 7 at x = −2
y = x2 + 6x + 7
y'=2x+6
y'=2(-2) + 6
What is 2
It is the Anti Derivative of:
sec2(x)
What is tan(x) + c
It is the evaluation of the definite integral.
y = x2/2 − 2x − 1; [−1, 1]
What is 0
It is the area under the curve over the given interval.
y=sec2(x) ; [-pi , -3pi/4]
I-pi-3pi/4 sec2(x) dx
What is 1
It is the U value of the substitution.
I0-1 18x2(3x2+3)2 dx
What is u=3x2+3
It is the equation of the line tangent to the function at the given point. Your answer should be in slope-intercept form
y = x3 − 3x2 + 2 at (3, 2)
y = x3 − 3x2 + 2
y'=3x2- 6x
y'=3(3)2-6(3) =9 y=9x+b
(2)=9(3)+b
2=27+b b=-25 What is y=9x-25
It is the Anti Derivative of:
1/(1-x2)1/2
What is sin-1(x) + c
It is the evaluation of the definite integral.
y = x3 + 3x2 − 2; [−2, 0]
What is { (−3 + (3)1/2)/3 , (−3 − (3)1/2)/3 }
It is the area under the curve over the given interval.
y=(x)1/2
I54 (x)1/2 dx
What is 2(5(5)1/2-8) / 3 = 2.12
It is the derivative of the U value of the substitution.
I20 2x[x2+4]2dx
U=x2+4
du=2x dx
What is du= 2x dx
It is the values of c that satisfy the Mean Value Theorem
y = −x2 + 8x − 17 ; [3, 6]
y = −x2 + 8x − 17
y'= -2x + 8
-1 = -2c + 8
What is 9/2
It is the Anti Derivative of:
sec(x)
What is ln|sec(x)+tan(x)| + c
It is the evaluation of the definite integral.
y = −x3 + 4x2 − 3; [0, 4]
What is 8/3
It is the area under the curve over the given interval.
y=-cos(x) ; [3pi/4 , pi]
Ipi3pi/4 -cos(x) dx
(2)1/2/2
What is 0.707
It is the integral of this equation.
I10 16x/(4x2+4)2 dx ; u=4x2+4
What is 1/4=.25
It is the slope of the function at the given value.
y = −ln (−x + 2) at x = −3
y = −ln (−x + 2)
y'= 1/-x+2
y'= 1/-(-3)+2
What is 1/5
It is the Anti Derivative of:
1/(a2+x2)
What is 1/a tan-1(x/a) + c
It is the evaluation of the definite integral.
y = −(−5x + 25)1/2 ; [3, 5]
What is 9/2
It is the area under the curve over the given interval.
y=6x2+6x ; [-.5 , .5]
What is 0
It is the integral of this expression.
I (3x+2)5 dx
(3x+2)6/18 +c