# Homework Help

eNotes Homework Help is a way for educators to help students understand their school work. Our experts are here to answer your toughest academic questions! Once it’s posted to our site, your question could help thousands of other students.

### Showing All Questions Answered Popular Recommended Unanswered Editor's Choice in All Subjects Literature History Science Math Arts Business Social Sciences Law and Politics Health Religion Other

• 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

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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 =...

Asked by enotes on via web

• 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 =...

Asked by enotes on via web

• 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 =...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• Social Sciences
Question 1: The best answer is D. All of these groups or individuals can have some influence over the economy of the United States. The greatest influence on the economy is held by the people....

• 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:...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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` `` ` `

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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` `:.`...

Asked by enotes on via web

• 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,`...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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.`...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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` `...

Asked by enotes on via web

• 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=...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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.`

Asked by enotes on via web

• 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.`

Asked by enotes on via web

• 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 +...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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`...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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 =...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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:...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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 -...

Asked by enotes on via web

• 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`...

Asked by enotes on via web