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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

MathYou 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...

MathYou 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...

MathYou 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 =...

MathYou 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 =...

MathYou 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 =...

MathYou 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...

MathYou 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...

Social SciencesQuestion 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....

MathYou 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:...

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`...

Math`a(t)=t^24t+6` `a(t)=v'(t)` `v(t)=inta(t)dt` `v(t)=int(t^24t+6)dt` `v(t)=t^3/34(t^2/2)+6t+c_1` `v(t)=t^3/32t^2+6t+c_1` `s(t)=intv(t)dt` `s(t)=int(t^3/3)2t^2+6t+c_1)dt`...

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` `` ` `

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`...

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=10+C` `C=1` `:.`...

MathAs 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,`...

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+t2`...

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`...

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`...

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`...

MathIntegrate 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.`...

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^2cos(x)+C_1x+C_2` Now lets find constants C_1 and C_2 , given...

Math`f''(t)=2e^t+3sin(t)` `f'(t)=int(2e^t+3sin(t))dt` `f'(t)=2e^t3cos(t)+C_1` `f(t)=int(2e^t3cos(t)+C_1)dt` `f(t)=2e^t3sin(t)+C_1t+C_2` Now let's find constants C_1 and C_2 , given f(0)=0 and...

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`...

MathYou 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...

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` `...

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=...

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...

MathYou 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...

MathThe 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.`

MathIntegrating gives `f(t)=t^2/2(1/2)t^(2)+C,` where C is any constant. `f(1)=1/21/2+C=C=6,` so C=6. The answer: `f(t)=t^2/21/(2t^2)+6.`

MathYou 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 +...

MathYou need to evaluate the antiderivative of the function f'(x), such that: `int f'(x)dx = f(x) + c` `int (x^21)/x dx = int x^2/x dx  int 1/x dx` `int (x^21)/x dx = int x dx  int 1/x dx` `int...

MathYou 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...

Math`f'(x)=4/sqrt(1x^2)` `f(x)=int(4/sqrt(1x^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`...

Math`f''(x)=2+12x12x^2` `f'(x)=int(2+12x12x^2)dx` `f'(x)=2x+12(x^2/2)12(x^3/3)+c_1` `f'(x)=2x+6x^24x^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`...

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`...

Math`f(x) =1/52/x` To determine the most general antiderivative of this function, take the integral of it. `F(x)=int f(x)dx = int(1/52/x)dx` Then, apply the integral formula `int cdu=cu +C` and `int...

MathThe 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...

MathThe 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...

MathThe 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 =...

MathThe 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...

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...

MathIntegrating f gives (both integrals are wellknown) `5e^x3sinh(x)+C,` where `C` is any constant. This is the answer. Checking by differentiation:...

MathYou 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...

MathYou 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...

MathYou 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...

MathYou 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 ...

Math`f(x)=43(1+x^2)^1` `F=int(43(1+x^2)^1)dx` `F=int4dxint3/(1+x^2)dx` `F=4x3arctan(x)+c_1` Let's find constant c_1 , given F(1)=0 `F(1)=0=4(1)3arctan(1)+c_1` `0=43(pi/4)+c_1` `c_1=(3pi)/44`...

Math`f''(x)=20x^312x^2+6x` `f'(x)=int(20x^312x^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^44x^3+3x^2+c_1` `f(x)=intf'(x)dx`...