Math Homework Help

• Math
You need to evaluate the indefinite integral, such that: `int sec y(tan y - sec y) dy= int (sin y - 1)/(cos^2 y) dy = int (sin y)/(cos^2 y) dy - int 1/(cos^2 y) dy` You need to solve `int (sin...

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You need to evaluate the indefinite integral and yo need to use the following trigonometric formula `1 + tan^2 y = 1/(cos^2 y).` Replacing `1/(cos^2 y)` for `1 + tan^2 y` yields: `int (1 + tan^2 y)...

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• Math
You need to evaluate the indefinite integral, such that: `int(4x - csc^2 x) dx= int 4x dx- int csc^2 x dx` `int(4x - csc^2 x) dx= int 4x dx- int 1/(sin^2 x) dx` You need to remember that `1/(sin^2...

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• Math
To find the solution of the differential equation `f'(x)=6x` ` ` with initial condition `f(0)=8 ` , integrate both sides: `int(f'(x) dx)=int(6x dx) =>f(x)+c_1=3x^2+c_2` ,where `c_1,c_2 ` are...

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• Math
You need to use direct integration to evaluate the general solution to the differential equation: `int (4x^2)dx = 4x^3/3 + c` You need to find the particular solution using the information...

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• Math
You need to use direct integration to evaluate the general solution to the differential equation: `int (8t^3 + 5)dt = int 8t^3 dt + int 5dt` `int (8t^3 + 5)dt = 8t^4/4 + 5t + c` `int (8t^3 + 5)dt =...

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• Math
You need to use direct integration to evaluate the general solution to the differential equation: `int (10s - 12s^3)ds = int 10s ds - int 12s^3 ds` `int (10s - 12s^3)ds = (10/2)s^2 - (12/4)s^4 +...

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• Math
`f''(x)=2` `f'(x)=int2dx` `f'(x)=2x+C_1` Now let's find constant C_1 , given f'(2)=5 `5=2(2)+C_1` `5=4+C_1` `C_1=1` `:.f'(x)=2x+1` `f(x)=int(2x+1)dx` `f(x)=(2x^2)/2+x+C_2` `f(x)=x^2+x+C_2` Now...

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• Math
`f''(x)=x^2` `f'(x)=intx^2dx` `f'(x)=x^3/3+C_1` Now let's find C_1 , given f'(0)=8 `8=0^3/3+C_1` `C_1=8` `:.f'(x)=x^3/3+8` `f(x)=int(x^3/3+8)dx` `f(x)=(1/3)((x^(3+1))/(3+1))+8x+C_2`...

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• Math
`int(sqrt(x)+1/(2sqrt(x)))dx` apply the sum rule `=intsqrt(x)dx+int1/(2sqrt(x))dx` Now, `intsqrt(x)dx` apply the power rule `=x^(1/2+1)/(1/2+1)` `=2/3x^(3/2)` `int1/(2sqrt(x))dx`...

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• Math
You need to find the indefinite integral, hence you need to use the following formula: `int x^n dx = (x^(n+1))/(n+1) + c` Replacing `2/3` for n yields: `int x^(2/3) dx = (x^(2/3+1))/(2/3+1) + c`...

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• Math
You need to evaluate the indefinite integral, such that: `int (root(4)(x^3) + 1) dx = int root(4)(x^3) dx + int dx` You may use the following formula `int x^n dx = (x^(n+1))/(n+1) + c` `int...

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• Math
You need to find the indefinite integral, hence you need to use the following formula: `int 1/(x^n) dx = (x^(-n+1))/(1-n)` Replacing 5 for n yields: `int 1/(x^5) dx = (x^(-5+1))/(1-5)` `int 1/(x^5)...

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• Math
You need to find the indefinite integral, hence you need to use the following formula: `int 1/(x^n) dx = (x^(-n+1))/(1-n)` Replacing 7 for n yields: `int 3/(x^7) dx = (3x^(-7+1))/(1-7)` `int...

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• Math
`int((x+6)/sqrt(x))dx` apply the sum rule `=int(x/sqrt(x))dx+int(6/sqrt(x))dx` Now, `int(x/sqrt(x))dx=intsqrt(x)dx` apply the power rule `=x^(1/2+1)/(1/2+1)` `=2/3x^(3/2)`...

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• Math
`int((x^4-3x^2+5)/x^4)dx` apply the sum rule `=intx^4/x^4dx-int(3x^2)/x^4dx+int5/x^4dx` `=int1dx-int3x^-2dx+int5x^-4dx` `=x-3(x^(-2+1)/(-2+1))+5(x^(-4+1)/(-4+1))` `=x+3x^-1+5(x^-3)/(-3)`...

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• Math
You need to evaluate the indefinite integral, hence, you need to open the brackets such that: `(x + 1)(3x - 2) = 3x^2 - 2x + 3x - 2 = 3x^2 + x - 2` `int(x + 1)(3x - 2)dx = int (3x^2 + x - 2)dx` You...

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• Math
`int(4t^2+3)^2dt` `=int(16t^4+24t^2+9)dt` apply the sum rule and power rule, `=int16t^4dt+int24t^2dt+int9dt` `=16(t^(4+1)/(4+1))+24(t^(2+1)/(2+1))+9t` `=(16t^5)/5+8t^3+9t+C` C is constant

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• Math
You need to evaluate the indefinite integral, hence, you need to split the integral, such that: `int (5cos x + 4sin x) dx = int 5cos x*dx + int 4sin x*dx` `int 5cos x*dx = 5sin x + c` `int 4sin...

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• Math
You need to evaluate the indefinite integral, hence, you need to split the integral, such that: `int (t^2 - cos t) dt = int t^2 dt - int cos t dt` You need to use the following formula `int t^n dt...

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• Math
You need to evaluate the indefinite integral, hence, you need to split the integral, such that: `int (x + 7) dx = int x dx - int 7 dx` You need to use the following formula `int x^n dx =...

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• Math
You need to evaluate the indefinite integral, hence, you need to split the integral, such that: `int (13 - x) dx = int 13 dx - int x dx` You need to use the following formula `int x^n dx =...

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• Math
You need to evaluate the indefinite integral, hence, you need to split the integral, such that: `int (x^5 + 1) dx = int x^5 dx + int dx` You need to use the following formula` int x^n dx =...

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• Math
You need to evaluate the indefinite integral, hence, you need to split the integral, such that: `int (8x^3 - 9x^2 + 4) dx = int 8x^3 dx - int 9x^2 dx + int 4dx` You need to use the following...

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• Math
You need to evaluate the indefinite integral, hence, you need to split the integral, such that: `int (x^(3/2) + 2x + 1) dx = int x^(3/2) dx + int 2x dx + int dx` You need to use the following...

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• Math
You need to remember the relation between acceleration, velocity and position, such that: `int a(t)dt = v(t) + c` `int v(t) dx = s(t) + c` You need to find first the velocity function, such that:...

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• Math
`a(t)=10sin(t)+3cos(t)` `a(t)=v'(t)` `v(t)=inta(t)dt` `v(t)=int(10sin(t)+3cos(t))dt` `v(t)=-10cos(t)+3sin(t)+c_1` `v(t)=s'(t)` `s(t)=intv(t)dt` `s(t)=int(-10cos(t)+3sin(t)+c_1)dt`...

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• Math
`a(t)=t^2-4t+6` `a(t)=v'(t)` `v(t)=inta(t)dt` `v(t)=int(t^2-4t+6)dt` `v(t)=t^3/3-4(t^2/2)+6t+c_1` `v(t)=t^3/3-2t^2+6t+c_1` `s(t)=intv(t)dt` `s(t)=int(t^3/3)-2t^2+6t+c_1)dt`...

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• Math
`f''(x) = x^-2` `f'(x) = -x^-1 + a` `f(x) = -lnx + ax + b` `Now, f(1) = f(2) = 0` `thus, a + b = 0` `and , -ln2 + 2a + b = 0` `thus, a = ln2 ` `b = -ln2` `thus, f(x) = -lnx + xln2 - ln2` `` ` `

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• Math
`f'''(x)=cos(x)` `f''(x)=intcos(x)dx` `f''(x)=sin(x)+C_1` Now let's find C_1 , given f''(0)=3 `f''(0)=3=sin(0)+C_1` `3=0+C_1` `C_1=3` `:.f''(x)=sin(x)+3` `f'(x)=int(sin(x)+3)dx`...

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• Math
`v(t)=sin(t)-cos(t)` `v(t)=s'(t)` `s(t)=intv(t)dt` `s(t)=int(sin(t)-cos(t))dt` `s(t)=-cos(t)-sin(t)+C` Let's find the constant C , given s(0)=0 `s(0)=0=-cos(0)-sin(0)+C` `0=-1-0+C` `C=1` `:.`...

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• Math
As I understand we have to find s(t). `s(t)=int(v(t))dt=int(1.5sqrt(t))dt=1.5*(2/3)*t^(3/2)+C=t^(3/2)+C,` where C is any constant. Also is given that s(4)=10, so `4^(3/2)+C=10,`...

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• Math
`a(t)=2t+1` `a(t)=v'(t)` `v(t)=inta(t)dt` `v(t)=int(2t+1)dt` `v(t)=2(t^2/2)+t+c_1` `v(t)=t^2+t+c_1` Now let's find constant c_1 given v(0)=-2 `v(0)=-2=0^2+0+c_1` `c_1=-2` `:.v(t)=t^2+t-2`...

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• Math
`f''(theta)=sin(theta)+cos(theta)` `f'(theta)=int(sin(theta)+cos(theta))d(theta)` `f'(theta)=-cos(theta)+sin(theta)+C_1` Now let's find constant C_1 , given f'(0)=4 `f'(0)=4=-cos(0)+sin(0)+C_1`...

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• Math
`f''(t)=3/sqrt(t)` `f'(t)=int(3/sqrt(t))dt` `f'(t)=3(t^(-1/2+1)/(-1/2+1))+C_1` `f'(t)=6sqrt(t)+C_1` Now let's find constant C_1 given f'(4)=7, `f'(4)=7=6sqrt(4)+C_1` `7=12+C_1` `C_1=-5`...

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• Math
`f''(x)=4+6x+24x^2` `f'(x)=int(4+6x+24x^2)dx` `f'(x)=4x+(6x^2)/2+(24x^3)/3+C_1` `f'(x)=4x+3x^2+8x^3+C_1` `f(x)=int(4x+3x^2+8x^3+C_1)dx` `f(x)=(4x^2)/2+(3x^3)/3+(8x^4)/4+C_1(x)+C_2`...

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• Math
Integrate this once (all integrals are the table ones): `f'(x)=(1/4)x^4+cosh(x)+C_1` and twice: `f(x)=(1/20)x^5+sinh(x)+C_1*x+C_2.` Now determine the constants `C_1` and `C_2` : `f(0)=C_2=1.`...

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• Math
`f''(x)=2+cos(x)` `f'(x)=int(2+cos(x)dx` `f'(x)=2x+sin(x)+C_1` `f(x)=int(2x+sin(x)+C_1)dx` `f(x)=2(x^2/2)-cos(x)+C_1x+C_2` `f(x)=x^2-cos(x)+C_1x+C_2` Now lets find constants C_1 and C_2 , given...

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• Math
`f''(t)=2e^t+3sin(t)` `f'(t)=int(2e^t+3sin(t))dt` `f'(t)=2e^t-3cos(t)+C_1` `f(t)=int(2e^t-3cos(t)+C_1)dt` `f(t)=2e^t-3sin(t)+C_1t+C_2` Now let's find constants C_1 and C_2 , given f(0)=0 and...

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• Math
`f''(x)=2/3x^(2/3)` `f'(x)=int2/3x^(2/3)dx` `f'(x)=2/3(x^(2/3+1)/(2/3+1))+c_1` `f'(x)=(2/3)(3/5)x^(5/3)+c_1` `f'(x)=2/5x^(5/3)+c_1` c_1 is constant `f(x)=int(2/5x^(5/3)+c_1)dx`...

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• Math
You need to evaluate f, knowing the second derivative, hence, you need to use the following principle, such that: `int f''(x)dx = f'(x) + c` `int f'(x) dx = f(x) + c` Hence, you need to evaluate...

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• Math
`f'''(t)=cos(t)` `f''(t)=intf'''(t)dt` `f''(t)=intcos(t)dt` `f''(t)=sin(t)+c_1` `f'(t)=int(sin(t)+c_1)dt` `f'(t)=-cos(t)+c_1t+c_2` `f(t)=intf'(t)dt` `f(t)=int(-cos(t)+c_1t+c_2)dt` `...

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`f'''(t)=e^t+t^(-4).` Integrate this once: `f''(t)=e^t+(-1/3)t^(-3)+C_1,` twice: `f'(t)=e^t+(-1/2)(-1/3)t^(-2)+C_1t+C_2,` thrice: `f(t)=e^t+(-1/1)(-1/2)(-1/3)t^(-1)+C_1t^2+C_2t+C_3=...

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• Math
`f'(x)=1+3sqrt(x)` `f(x)=int(1+3sqrt(x))dx` `f(x)=x+3(x^(1/2+1)/(1/2+1))+c` `f(x)=x+3(x^(3/2)/(3/2))+c` `f(x)=x+3(2/3)x^(3/2)+c` `f(x)=x+2x^(3/2)+c` Now let's find constant c , given f(4)=25...

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You need to evaluate f and the problem provides f'(x), hence, you need to use the following relation, such that: `int f'(x)dx = f(x)+ c` `int (5x^4 - 3x^2 + 4)dx = f(x) + c` You need to evaluate...

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The general antiderivative of f' is `f(t)=4arctan(t)+C,` where C is any constant. It must be f(1)=0, so C=-4arctan(1). acrctan(1) is `pi/4,` so the answer is `f(t)=4arctan(t)-pi.`

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Integrating gives `f(t)=t^2/2-(1/2)t^(-2)+C,` where C is any constant. `f(1)=1/2-1/2+C=C=6,` so C=6. The answer: `f(t)=t^2/2-1/(2t^2)+6.`

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You need to evaluate the antiderivative of the function f'(t), such that: `int f'(t)dt = f(t) + c` `int (2cos t + sec^2 t) dt = int 2cos t dt + int sec^2 t dt` `int (2cos t + sec^2 t) dt = 2sin t +...

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• Math
You need to evaluate the antiderivative of the function f'(x), such that: `int f'(x)dx = f(x) + c` `int (x^2-1)/x dx = int x^2/x dx - int 1/x dx` `int (x^2-1)/x dx = int x dx - int 1/x dx` `int...

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You need to evaluate f and the problem provides f'(x), hence, you need to use the following relation, such that: `int f'(x)dx = f(x)+ c` You need to evaluate the indefinite integral of the power...

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