# Math Homework Help

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• Math
We have to be careful here because of absolute value sign. We need to know where the function under absolute value changes sign (in the interval of integration), which is at x=pi in this case....

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• Math
Hello! Find the indefinite integral: int((sqrt(y)-y)/y^2)dy=int(y^(-3/2)-1/y)dy=(-2)*y^(-(1)/(2))-ln(y). Then substitute y from 1 to 4: (-2*4^(-1/2)-ln(4))-(-2)=-1-ln(4)+2=1-ln(4) approx -0.386.

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• Math
Hello! First find the indefinite integral, int(x/(2)-2/x)dx=int(x/2)dx-int(2/x)dx=(x^2)/4-2ln(x). Then substitute x from 1 to 2: (2^2/4-2ln(2))-(1^2/4-2ln(1))=1-2ln(2)-1/4=3/4-2ln(2) approx...

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• Math
You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_0^1(5x - 5^x)dx = int_0^1(5x)dx - int_0^1 5^x dx...

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• Math
You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_0^1(x^10 + 10^x)dx = int_0^1(x^10)dx + int_0^1 10^x...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(u) du = F(b) - F(a) int_(pi/4)^(pi/3) csc^2 theta d theta = int_(pi/4)^(pi/3) 1/(sin^2...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_0^(pi/4)(1+cos^2 theta)/(cos^2 theta) d theta =...

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• Math
Before evaluating this integral, simplify the expression in the integral using trigonometric identities. The following Pythagorean identity will be useful: tan^2(theta) + 1 = sec^2(theta) Start...

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• Math
We will make substitution x=t^6. Therefore, the differential is dx=6t^5dt and the new bounds of integration are t_1=root(6)(1)=1 and t_2=root(6)(64)=2. int_1^64(1+root(3)(x))/sqrt x...

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• Math
Hello! Consider the denominator first: sinh(x)+cosh(x)=(e^x-e^(-x))/(2) +(e^x+e^(-x))/(2)=e^x. Therefore the integrand is equal to 2 and the integral is 2*(10-(-10))=40.

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• Math
Hello! This integral is a table one, int(dr)/(sqrt(1-r^2))=arcsin(r)+C.   Therefore the definite integral is equal to arcsin(sqrt(3)/2)-arcsin(0)=pi/3-0=pi/3 approx 1.047. =arcsin(r)+C.

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) You need to expand the cube such that: (x-1)^3 = x^3 - 1 - 3x(x-1)...

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• Math
Hello! (t^2-1)/(t^4-1)=1/(t^2+1). Therefore indefinite integral is arctan(t)(+C). Substitute t from 0 to 1/sqrt(3) and obtain arctan(1/sqrt(3))-arctan(0)=pi/6-0=pi/6 approx 0.524.

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• Math
We have to be careful because of the absolute value. We need to know where the function under absolute values is positive and when it is negative in the given interval. It is also a good idea to...

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• Math
Hello! Consider two intervals: (-1, 0) and (0, 2) to simplify |x|: int_(-1)^2(x-2|x|)dx=int_(-1)^0(x-2|x|)dx+int_0^2(x-2|x|)dx= =int_(-1)^0(x+2x)dx+int_0^2(x-2x)dx=...

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• Math
You need to find the indefinite integral, hence, you need to remember that sec t = 1/(cos t) , such that: int (sec t)(sec t + tan t) dt = int (1/(cos t))(1 + sin t)/(cos t) dt int (sec t)(sec...

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• Math
You need to find the indefinite integral, hence, you need to remember that 1 + tan^2 x = 1/(cos^2 x) = (tan x)'. int (1 + tan^2 x )dx = int (tan x)' dx = tan x + c Hence, evaluating the...

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• Math
You need to find the indefinite integral, hence, you need to remember that sin 2x = 2sin x*cos x , such that: int (sin 2x)(sin x) dx = int (2sin x*cos x)/(sin x) dx  int (sin 2x)(sin x) dx =...

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• Math
You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_(-2)^3(x^3 - 3)dx = int_(-2)^3(x^3)dx - int_(-2)^3 3...

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• Math
You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_1^2(4x^3 - 3x^2 + 2x)dx = int_1^2(4x^3)dx - int_1^2...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t)dt = int_(-2)^0 ((1/2)t^4 dt +...

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• Math
We will use linearity of integral: int(a cdot f(x)+b cdot g(x))dx=a int f(x)dx+b int g(x)dx. int_0^3(1+6w^2-10w^4)dw=int_0^3dw+6int_0^3w^2dw-10int_0^3w^4dw=...

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• Math
We have to evaluate the integral: \int_{0}^{2}(2x-3)(4x^2+1)dx=\int_{0}^{2}(8x^3-12x^2+2x-3)dx =[8(x^4/4)-12(x^3/3)+2(x^2/2)-3x]_{0}^{2}...

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• Math
You need to use the following substitution to evaluate the definite integral, such that: 1 - t = u => -dt = du int_(-1)^1 t*(1-t)^2dt = int_(u_1)^(u_2) (1 - u)*u^2 (-du) int_(u_1)^(u_2) (u...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_0^pi (5e^x+ 3sin x)dx = int_0^pi 5e^x dx + int_0^pi 3sin x...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_1^2 (1/x^2 - 4/x^3)dx = int_1^2 1/x^2 dx - int_1^2 4/x^3 dx...

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• Math
int_1^4(4+6u)/sqrt(u)du =int_1^4(4/sqrt(u)+(6u)/sqrt(u))du =int_1^4(4u^(-1/2)+6u^(1/2))du =[4(u^(-1/2+1)/(-1/2+1))+6(u^(1/2+1)/(1/2+1))]_1^4 =[4(u^(1/2)/(1/2))+6(u^(3/2)/(3/2))]_1^4...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_0^4(3sqrt t - 2e^t)dt = int_0^4 3sqrt tdt - int_0^4 2e^t...

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• Math
You need to evaluate the definite integral, such that: int_0^1 x(root(3) x + root(4) x)dx = int_0^1 (x^(1+1/3) + x^(1+1/4))dx int_0^1 x(root(3) x + root(4) x)dx = ((x^(2+1/3))/(2+1/3) +...

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• Math
You need to evaluate the indefinite integral, such that: int f(x)dx = F(x) + c int (x^2 - x^(-2))dx = int (x^2)dx - int x^(-2) dx  Evaluating each definite integral, using the formula int x^n...

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• Math
int sqrt(x^3)root(3)(x^2) dx Before evaluating, convert the radicals to expressions with rational exponents. = int x^(3/2)*x^(2/3) dx Then, simplify the integrand. Apply the laws of exponent...

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• Math
int (x^4-1/2x^3+1/4x-2)dx To evaluate this integral, apply the formulas int x^n dx=x^(n+1)/(n+1) +C and int adx = ax + C . int (x^4-1/2x^3+1/4x-2)dx =x^5/5 - 1/2*x^4/4 + 1/4*x^2/2-2x...

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• Math
int (y^3+1.8y^2-2.4y)dy To evaluate this integral, apply the formula int x^n dx = x^(n+1)/(n+1) + C . = y^4/4 + 1.8y^3/3 - 2.4y^2/2 + C =0.25y^4 + 0.6y^3 - 1.2y^2 + C Therefore, ...

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• Math
You need to evaluate the indefinite integral, hence, you need to open the brackets, such that: (u+4)(2u+1) = 2u^2 + 9u + 4 int (u+4)(2u+1) du = int (2u^2 + 9u + 4) du  int (u+4)(2u+1) du =...

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• Math
intv(v^2+2)^2dv =intv((v^2)^2+2v^2*2+2^2)dv =intv(v^4+4v^2+4)dv =int(v^5+4v^3+4v)dv apply the power rule, =v^6/6+4v^4/4+4v^2/2+C , C is constant =v^6/6+v^4+2v^2+C

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• Math
int(x^3-2sqrt(x))/xdx Simplify by dividing each term in the numerator by x.  <br>  =(x^3/3)-(2x^(1/2))/(1/2)+C =(x^3/3)-4x^(1/2)+C   

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• Math
int(x^2+1+1/(x^2+1))dx apply the sum rule, =intx^2dx+int1dx+int1/(x^2+1)dx To evaluate the above integrals, we know that, intx^ndx=x^(n+1)/(n+1) and int1/(x^2+1)dx=arctan(x) using above,...

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• Math
We have to find the integral : \int (sin(x)+sinh(x) )dx=-cos(x)+cosh(x)+C where C is a constant.

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• Math
int(csc^2(t)-2e^t)dt Apply the sum rule, =intcsc^2(t)dt-int(2e^tdt We now the following common integrals, intcsc^2(x)=-cot(x) and inte^xdx=e^x evaluate using the above, =-cot(t)-2e^t+C...

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• Math
You need to evaluate the indefinite integral, such that: int f(theta)d theta = F(theta) + c int (theta - csc theta* cot theta)d theta = int theta d theta - int (csc theta* cot theta)d theta...

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• Math
y=sec^2(x) Refer the graph in the attached image. From the graph, Area of the region beneath the curve ~~ 4/10(Area of the Rectangle) Area of the region=~~(4/10)(pi/3*4) Area of the...

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• Math
We have to evaluate the integral and interpret as difference of areas. So, \int_{-1}^{2}x^3dx=[x^4/4]_{-1}^{2} =[(2^4)/4]-1/4 =16/4-1/4 =15/4=3.75 So...

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• Math
The integral actually represents difference between red area and green area shown in the image below i.e. int_(pi/6)^(2pi)cos(x)dx=P_1-P_2+P_3...

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• Math
g(x)=int_(2x)^(3x)(u^2-1)/(u^2+1)du g(x)=int_(2x)^0(u^2-1)/(u^2+1)du+int_0^(3x)(u^2-1)/(u^2+1)du g(x)=-int_0^(2x)(u^2-1)/(u^2+1)du+int_0^(3x)(u^2-1)/(u^2+1)du...

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• Math
Hello! Let's temporarily denote the antiderivative of tsin(t) as A(t). Then the integral is equal to A(1+2x)-A(1-2x) and its derivative is A'(1+2x)*2-A'(1-2x)*(-2)=2*(A'(1+2x)+A'(1-2x))....

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• Math
F(x)=int_x^(x^2)(e^(t^2))dt From the fundamental theorem of calculus, int_x^(x^2)(e^(t^2))dt=F(x^2)-F(x) d/dxint_x^(x^2)(e^(t^2))dt=F'(x^2).d/dx(x^2)-F'(x) =2x(e^((x^2)^2))-e^(x^2)...

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• Math
Hello! Let's temporarily denote the antiderivative of arctan(t) as A(t). Then F(x)= A(2x)-A(sqrt(x)), and F'(x)=A'(2x)*2-A'(sqrt(x))*(1/(2sqrt(x))). Recall what A'(x) is and obtain...

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• Math
You need to evaluate the the derivative of the function, hence, you need to use the part1 of fundamental theorem of calculus: y = int_a^b f(x)dx => (dy)/(dx) = f(x) for x in (a,b) If f(x) =...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(u) du = F(b) - F(a) int_0^3 (2sin x - e^x) dx =int_0^3 2sin x dx - int_0^3 e^x dx...

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• Math
int_1^2((v^3+3v^6)/v^4)dv simplify the integrand and apply the sum rule, =int_1^2(v^3/v^4+(3v^6)/v^4)dv =int_1^2(1/v+3v^2)dv using the following common integrals int1/xdx=ln(x) and...

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