# Math Homework Help

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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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• Math
`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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• Math
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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• Math
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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• Math
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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• Math
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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• Math
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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• Math
`f'(x)=4/sqrt(1-x^2)` `f(x)=int(4/sqrt(1-x^2))dx` `f(x)=4arcsin(x)+c_1` Let's find constant c_1 , given f(1/2)=1 `f(1/2)=1=4arcsin(1/2)+c_1` `1=4(pi/6)+c_1` `c_1=1-(2pi)/3`...

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• Math
`f''(x)=-2+12x-12x^2` `f'(x)=int(-2+12x-12x^2)dx` `f'(x)=-2x+12(x^2/2)-12(x^3/3)+c_1` `f'(x)=-2x+6x^2-4x^3+c_1` Now let's find constant c_1 , given f'(0)=12 `f'(0)=12=-2(0)+6(0^2)-4(0^3)+c_1`...

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• Math
`f''(x)=8x^3+5` `f'(x)=int(8x^3+5)dx` `f'(x)=8(x^4/4)+5x+c_1` `f'(x)=2x^4+5x+c_1` Now let's find constant c_1 , given f'(1)=8 `f'(1)=8=2(1)^4+5(1)+c_1` `8=2+5+c_1` `c_1=1` `:.f'(x)=2x^4+5x+1`...

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• Math
`f(x) =1/5-2/x` To determine the most general antiderivative of this function, take the integral of it. `F(x)=int f(x)dx = int(1/5-2/x)dx` Then, apply the integral formula `int cdu=cu +C` and `int...

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• Math
The most general antiderivative F(t) of the function f(t) can be found using the following relation: `int f(t)dt = F(t) + c` `int (3t^4 - t^3 + 6t^2)/(t^4)dt = int (3t^4)/(t^4)dt - int...

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• Math
The most general antiderivative G(t) of the function g(t) can be found using the following relation: `int g(t)dt = G(t) + c` `int (1 + t + t^2)/(sqrt t)dt = int (1)/(sqrt t)dt + int (t)/(sqrt t)dt...

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• Math
The most general antiderivative `R(theta)` of the function `r(theta)` can be found using the following relation: `int r(theta)d theta = R(theta) + c` `int (sec theta*tan theta - 2 e^theta)d theta =...

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• Math
The most general antiderivative `H(theta)` of the function `h(theta)` can be found using the following relation: `int h(theta)d theta = H(theta) + c` `int (2sin theta - sec^2 theta)d theta = int...

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• Math
`int f(t) dt=int sin(t)+2 sinh(t) dt = int sin(t) dt + 2 int sinh(t) dt ` We note that the derivative of `cosh(t)=sinh(t) =>d/dx cosh(t) = sinh(t)=>d/dx sinh(t)=cosh(t)` This means that `int...

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• Math
Integrating f gives (both integrals are well-known) `5e^x-3sinh(x)+C,` where `C` is any constant. This is the answer. Checking by differentiation:...

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• Math
You need to evaluate the most general antiderivative, using the following rule, such that: `int f(x) dx = F(x) + c` `int (2sqrtx + 6cos x) dx = int 2sqrt x dx + int 6cos x dx` `int (2sqrtx + 6cos...

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• Math
You need to evaluate the most general antiderivative, using the following rule, such that: `int f(x) dx = F(x) + c` `int (x^5 - x^3 + 2x)/(x^4) dx = int(x^5)/(x^4) dx - int (x^3)/(x^4)dx + int...

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• Math
You need to evaluate the most general antiderivative, using the following rule, such that: `int f(x) dx = F(x) + c` `int (2 +x^2)/(1+x^2) dx = int(1+x^2)/(1+x^2) dx + int 1/(1+x^2)dx ` `int (2...

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• Math
You need to evaluate the antiderivative F,under the given condition, such that: `int f(x)dx = F(x) + c` Hence, you need to evaluate the indefinite integral of function f(x), such that: `int (5x^4 -...

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• Math
`f(x)=4-3(1+x^2)^-1` `F=int(4-3(1+x^2)^-1)dx` `F=int4dx-int3/(1+x^2)dx` `F=4x-3arctan(x)+c_1` Let's find constant c_1 , given F(1)=0 `F(1)=0=4(1)-3arctan(1)+c_1` `0=4-3(pi/4)+c_1` `c_1=(3pi)/4-4`...

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• Math
`f''(x)=20x^3-12x^2+6x` `f'(x)=int(20x^3-12x^2+6x)dx` Applying the sum rule and power yields, `f'(x)=20(x^4/4)-12x^3/3+6x^2/2+c_1` `f'(x)=5x^4-4x^3+3x^2+c_1` `f(x)=intf'(x)dx`...

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• Math
`f''(x)=x^6-4x^4+x+1` `f'(x)=intf''(x)dx` `f'(x)=int(x^6-4x^4+x+1)dx` `f'(x)=x^7/7-4(x^5/5)+x^2/2+x+c_1` `f'(x)=x^7/7-(4x^5)/5+x^2/2+x+c_1` `f(x)=intf'(x)dx`...

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• Math
`f(x) = x - 3` To find the general antiderivative of this function, we take its integral. `F(x) = int f(x) dx = int(x-3)dx` Take the integral of each term. `=intxdx - int3dx` For the first...

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• Math
You need to evaluate the most antiderivative of the function, such that: `int f(x) dx = F(x) + c` `int ((1/2)x^2 - 2x + 6) dx = int(1/2)x^2 dx - int 2xdx+ int 6 dx` `int ((1/2)x^2 - 2x + 6) dx =...

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• Math
`f(x)=1/2+3/4x^2-4/5x^3 ` To determine the most general antiderivative of this function, take the integral of f(x). `F(x)=intf(x) dx= int(1/2+3/4x^2-4/5x^3)dx` Then, apply the integral formula `int...

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• Math
You need to evaluate the most antiderivative of the function, such that: `int f(x) dx = F(x) + c` `int (8x^9 - 3x^6 + 12x^3) dx = int8x^9 dx - int 3x^6dx+ int12x^3 dx` `int (8x^9 - 3x^6 + 12x^3) dx...

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• Math
First, you need to open the brackets, such that: `f(x) = (x+1)(2x-1) = 2x^2 + x - 1` You need to evaluate the most antiderivative of the function, such that: `int f(x) dx = F(x) + c` `int (2x^2 + x...

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• Math
You need to evaluate the most general antiderivative, using the following rule, such that: `int f(x) dx = F(x) + c` First, you need to open the brackets, such that: `f(x) = x(2-x)^2 = x*(4 - 4x +...

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• Math
The most general antiderivative F(x) of the function f(x) can be found using the following relation: `int f(x)dx = F(x) + c` `int (7x^(2/5) + 8x^(-4/5))dx = int (7x^(2/5))dx + int (8x^(-4/5))dx`...

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• Math
The most general antiderivative F(x) of the function f(x) can be found using the following relation: `int f(x)dx = F(x) + c` `int (x^(3.4) - 2x^(sqrt2 - 1))dx = int (x^(3.4))dx - int (2x^(sqrt2 -...

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• Math
The answer is `F(x) = x*sqrt(2) + C,` where C is any constant. Let's check: `F'(x) = (x*sqrt(2) + C)' = x'*sqrt(2) + 0 = 1*sqrt(2) = sqrt(2),` QED.

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• Math
`e^2` isn't depend on x, so its antiderivative is `x*e^2.` The general antiderivative is `x*e^2+C,` where C is any constant. Let's check: `(x*e^2+C)'=x'*e^2+C'=1*e^2+0=e^2.`

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• Math
The most general antiderivative F(x) of the function f(x) can be found using the following relation: `int f(x)dx = F(x) + c` `int (3sqrt x - 2root(3) x)dx = int (3sqrt x)dx - int (2root(3) x)dx`...

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• Math
`f(x)= root(3)(x^2) + xsqrtx` To determine the most general antiderivative of this function, take the integral of f(x). `F(x) = intf(x) dx = int (root(3)(x^2)+xsqrtx)dx` Express the radicals in...

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• Math
Hello! At first, consider line graph. It gives the same information as the bar graph, but line graph combines data from both cities which makes comparing easier. From the line graph we see those...

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• Math

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• Math
There are an infinite number of values on the number line between `1/3` and ```1/4` .`` One way to find a value between is to write equivalent fractions using a common denominator. Because the...

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