• MATH

∫10−102exsinh(x)+cosh(x)dx∫-10102exsinh(x)+cosh(x)dx Evaluate the integral

Hello! Consider the denominator first: sinh(x)+cosh(x)=ex−e−x2+ex+e−x2=ex.sinh(x)+cosh(x)=ex-e-x2+ex+e-x2=ex. Therefore the integrand is equal to 22 and the integral is 2*(10-(-10))=40.

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• MATH

∫6411+x−−√3x−−√dx∫1641+x3xdx Evaluate the integral

We will make substitution x=t6.x=t6. Therefore, the differential is dx=6t5dtdx=6t5dt and the new bounds of integration are t1=1–√6=1t1=16=1 and t2=64−−√6=2.t2=646=2.

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• MATH

∫π30sin(θ)+sin(θ)tan2(θ)sec2(θ)dθ∫0π3sin(θ)+sin(θ)tan2(θ)sec2(θ)dθ Evaluate the integral

Before evaluating this integral, simplify the expression in the integral using trigonometric identities. The following Pythagorean identity will be useful: tan2(θ)+1=sec2(θ)tan2(θ)+1=sec2(θ) Start…

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• MATH

∫π401+cos2(θ)cos2(θ)dθ∫0π41+cos2(θ)cos2(θ)dθ Evaluate the integral

You need to evaluate the definite integral using the fundamental theorem of calculus, such that:∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a)

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• MATH

∫π3π4csc2(θ)dθ∫π4π3csc2(θ)dθ Evaluate the integral

You need to evaluate the definite integral using the fundamental theorem of calculus, such that: ∫baf(u)du=F(b)−F(a)∫abf(u)du=F(b)-F(a)

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• MATH

∫10(x10+10x)dx∫01(x10+10x)dx Evaluate the integral

You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a)

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• MATH

∫10(5x−5x)dx∫01(5x-5x)dx Evaluate the integral

You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a) ∫10(5x−5x)dx=∫10(5x)dx−∫105xdx∫01(5x-5x)dx=∫01(5x)dx-∫015xdx…

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• MATH

∫21(x2−2x)dx∫12(x2-2x)dx Evaluate the integral

Hello! First find the indefinite integral, ∫(x2−2x)dx=∫(x2)dx−∫(2x)dx=x24−2ln(x).∫(x2-2x)dx=∫(x2)dx-∫(2x)dx=x24-2ln(x).Then substitute x from 1 to 2:

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• MATH

∫41y√−yy2dy∫14y-yy2dy Evaluate the integral

Hello! Find the indefinite integral: ∫(y√−yy2)dy=∫(y−32−1y)dy=(−2)⋅y−12−ln(y).∫(y-yy2)dy=∫(y-32-1y)dy=(-2)⋅y-12-ln(y).Then substitute y from 1 to 4: (−2⋅4−12−ln(4))−(−2)=−1−ln(4)+2=1−ln(4)≈−0.386.(-2⋅4-12-ln(4))-(-2)=-1-ln(4)+2=1-ln(4)≈-0.386.

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• MATH

∫10x(x−−√3+x−−√4)dx∫01x(x3+x4)dx Evaluate the integral

You need to evaluate the definite integral, such that: ∫10x(x−−√3+x−−√4)dx=∫10(x1+13+x1+14)dx∫01x(x3+x4)dx=∫01(x1+13+x1+14)dx

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• MATH

∫40(3t√−2et)dt∫04(3t-2et)dt Evaluate the integral

You need to evaluate the definite integral using the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a)

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• MATH

∫41(4+6uu−−√)du∫14(4+6uu)du Evaluate the integral

∫414+6uu−−√du∫144+6uudu =∫41(4u−−√+6uu−−√)du=∫14(4u+6uu)du =∫41(4u−12+6u12)du=∫14(4u-12+6u12)du =[4(u−12+1−12+1)+6(u12+112+1)]41=[4(u-12+1-12+1)+6(u12+112+1)]14 =[4(u1212)+6(u3232)]41=[4(u1212)+6(u3232)]14…

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• MATH

∫21(1×2−4×3)dx∫12(1×2-4×3)dx Evaluate the integral

You need to evaluate the definite integral using the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a) ∫21(1×2−4×3)dx=∫211x2dx−∫214x3dx∫12(1×2-4×3)dx=∫121x2dx-∫124x3dx…

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• MATH

∫π0(5ex+3sin(x))dx∫0π(5ex+3sin(x))dx Evaluate the integral

You need to evaluate the definite integral using the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a)

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• MATH

∫1−1t(1−t)2dt∫-11t(1-t)2dt Evaluate the integral

You need to use the following substitution to evaluate the definite integral, such that: 1−t=u⇒−dt=du1-t=u⇒-dt=du ∫1−1t⋅(1−t)2dt=∫u2u1(1−u)⋅u2(−du)∫-11t⋅(1-t)2dt=∫u1u2(1-u)⋅u2(-du)

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• MATH

∫20(2x−3)(4×2+1)dx∫02(2x-3)(4×2+1)dx Evaluate the integral

We have to evaluate the integral: ∫20(2x−3)(4×2+1)dx=∫20(8×3−12×2+2x−3)dx∫02(2x-3)(4×2+1)dx=∫02(8×3-12×2+2x-3)dx =[8(x44)−12(x33)+2(x22)−3x]20=[8(x44)-12(x33)+2(x22)-3x]02…

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• MATH

∫30(1+6w2−10w4)dw∫03(1+6w2-10w4)dw Evaluate the integral

We will use linearity of integral: ∫(a⋅f(x)+b⋅g(x))dx=a∫f(x)dx+b∫g(x)dx.∫(a⋅f(x)+b⋅g(x))dx=a∫f(x)dx+b∫g(x)dx. ∫30(1+6w2−10w4)dw=∫30dw+6∫30w2dw−10∫30w4dw=∫03(1+6w2-10w4)dw=∫03dw+6∫03w2dw-10∫03w4dw=…

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• MATH

∫0−2((12)t4+(14)t3−t)dt∫-20((12)t4+(14)t3-t)dt Evaluate the integral

You need to evaluate the definite integral using the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a)

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• MATH

∫21(4×3−3×2+2x)dx∫12(4×3-3×2+2x)dx Evaluate the integral

You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a)

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• MATH

∫3−2(x3−3)dx∫-23(x3-3)dx Evaluate the integral

You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a)

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• MATH

∫sin(2x)sin(x)dx∫sin(2x)sin(x)dx Find the general indefinite integral.

You need to find the indefinite integral, hence, you need to remember that sin 2x=2sinx⋅cosx2x=2sinx⋅cosx , such that: ∫(sin2x)(sinx)dx=∫2sinx⋅cosxsinxdx∫(sin2x)(sinx)dx=∫2sinx⋅cosxsinxdx

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• MATH

∫(1+tan2(x))dx∫(1+tan2(x))dx Find the general indefinite integral.

You need to find the indefinite integral, hence, you need to remember that 1+tan2x=1cos2x=(tanx)’.1+tan2x=1cos2x=(tanx)′. ∫(1+tan2x)dx=∫(tanx)’dx=tanx+c∫(1+tan2x)dx=∫(tanx)′dx=tanx+c Hence, evaluating the…

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• MATH

∫sec(t)(sec(t)+tan(t))dt∫sec(t)(sec(t)+tan(t))dt Find the general indefinite integral.

You need to find the indefinite integral, hence, you need to remember that sect=1costsect=1cost , such that: ∫(sect)(sect+tant)dt=∫(1cost)1+sintcostdt∫(sect)(sect+tant)dt=∫(1cost)1+sintcostdt

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• MATH

∫(θ−csc(θ)cot(θ))dθ∫(θ-csc(θ)cot(θ))dθ Find the general indefinite integral.

You need to evaluate the indefinite integral, such that: ∫f(θ)dθ=F(θ)+c∫f(θ)dθ=F(θ)+c ∫(θ−cscθ⋅cotθ)dθ=∫θdθ−∫(cscθ⋅cotθ)dθ∫(θ-cscθ⋅cotθ)dθ=∫θdθ-∫(cscθ⋅cotθ)dθ…

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• MATH

∫(csc2(t)−2et)dt∫(csc2(t)-2et)dt Find the general indefinite integral.

∫(csc2(t)−2et)dt∫(csc2(t)-2et)dt Apply the sum rule, =∫csc2(t)dt−∫(2etdt=∫csc2(t)dt-∫(2etdt We now the following common integrals, ∫csc2(x)=−cot(x)∫csc2(x)=-cot(x) and ∫exdx=ex∫exdx=ex evaluate using the above, =−cot(t)−2et+C=-cot(t)-2et+C…

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• MATH

∫(sin(x)+sinh(x))dx∫(sin(x)+sinh(x))dx Find the general indefinite integral.

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

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• MATH

∫(x2+1+1×2+1)dx∫(x2+1+1×2+1)dx Find the general indefinite integral.

∫(x2+1+1×2+1)dx∫(x2+1+1×2+1)dx apply the sum rule, =∫x2dx+∫1dx+∫1×2+1dx=∫x2dx+∫1dx+∫1×2+1dx To evaluate the above integrals, we know that, ∫xndx=xn+1n+1∫xndx=xn+1n+1 and ∫1×2+1dx=arctan(x)∫1×2+1dx=arctan(x) using above,…

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• MATH

∫x3−2x−−√xdx∫x3-2xxdx Find the general indefinite integral.

∫x3−2x−−√xdx∫x3-2xxdx Simplify by dividing each term in the numerator by x. <br><br> =(x33)−2×1212+C=(x33)-2×1212+C =(x33)−4×12+C=(x33)-4×12+C

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• MATH

∫v(v2+2)2dv∫v(v2+2)2dv Find the general indefinite integral.

∫v(v2+2)2dv∫v(v2+2)2dv =∫v((v2)2+2v2⋅2+22)dv=∫v((v2)2+2v2⋅2+22)dv =∫v(v4+4v2+4)dv=∫v(v4+4v2+4)dv =∫(v5+4v3+4v)dv=∫(v5+4v3+4v)dv apply the power rule, =v66+4v44+4v22+C=v66+4v44+4v22+C , C is constant =v66+v4+2v2+C=v66+v4+2v2+C

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• MATH

∫(u+4)(2u+1)du∫(u+4)(2u+1)du Find the general indefinite integral.

You need to evaluate the indefinite integral, hence, you need to open the brackets, such that: (u+4)(2u+1)=2u2+9u+4(u+4)(2u+1)=2u2+9u+4 ∫(u+4)(2u+1)du=∫(2u2+9u+4)du∫(u+4)(2u+1)du=∫(2u2+9u+4)du

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• MATH

∫y3+1.8y2−2.4ydy∫y3+1.8y2-2.4ydy Find the general indefinite integral.

∫(y3+1.8y2−2.4y)dy∫(y3+1.8y2-2.4y)dy To evaluate this integral, apply the formula ∫xndx=xn+1n+1+C∫xndx=xn+1n+1+C . =y44+1.8y33−2.4y22+C=y44+1.8y33-2.4y22+C =0.25y4+0.6y3−1.2y2+C=0.25y4+0.6y3-1.2y2+C Therefore,

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• MATH

∫x4−(12)x3+(14)x−2dx∫x4-(12)x3+(14)x-2dx Find the general indefinite integral.

∫(x4−12×3+14x−2)dx∫(x4-12×3+14x-2)dx To evaluate this integral, apply the formulas ∫xndx=xn+1n+1+C∫xndx=xn+1n+1+C and ∫adx=ax+C∫adx=ax+C . ∫(x4−12×3+14x−2)dx∫(x4-12×3+14x-2)dx

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• MATH

∫x3−−√x2−−√3∫x3x23 Find the general indefinite integral.

∫x3−−√x2−−√3dx∫x3x23dx Before evaluating, convert the radicals to expressions with rational exponents. =∫x32⋅x23dx=∫x32⋅x23dx Then, simplify the integrand. Apply the laws of exponent…

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• MATH

∫(x2−x−2)dx∫(x2-x-2)dx Find the general indefinite integral.

You need to evaluate the indefinite integral, such that: ∫f(x)dx=F(x)+c∫f(x)dx=F(x)+c ∫(x2−x−2)dx=∫(x2)dx−∫x−2dx∫(x2-x-2)dx=∫(x2)dx-∫x-2dxEvaluating each definite integral, using the formula

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• MATH

y=∫sin(x)cos(x)(ln(1+2v))dvy=∫cos(x)sin(x)(ln(1+2v))dv Find the derivative of the function.

You need to evaluate the the derivative of the function, hence, you need to use the part1 of fundamental theorem of calculus: y=∫baf(x)dx⇒dydx=f(x)y=∫abf(x)dx⇒dydx=f(x) for x∈(a,b)x∈(a,b) If

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• MATH

F(x)=∫2xx√(arctan(t))dtF(x)=∫x2x(arctan(t))dt Find the derivative of the function.

Hello! Let’s temporarily denote the antiderivative of arctan(t)arctan(t) as A(t).A(t).Then F(x)=A(2x)−A(x−−√),F(x)=A(2x)-A(x), and F'(x)=A'(2x)⋅2−A'(x−−√)⋅(12x−−√).F′(x)=A′(2x)⋅2-A′(x)⋅(12x). Recall what A'(x)A′(x) is and obtain…

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• MATH

F(x)=∫x2x(et2)dtF(x)=∫xx2(et2)dt Find the derivative of the function.

F(x)=∫x2x(et2)dtF(x)=∫xx2(et2)dt From the fundamental theorem of calculus, ∫x2x(et2)dt=F(x2)−F(x)∫xx2(et2)dt=F(x2)-F(x) ddx∫x2x(et2)dt=F'(x2).ddx(x2)−F'(x)ddx∫xx2(et2)dt=F′(x2).ddx(x2)-F′(x) =2x(e(x2)2)−ex2=2x(e(x2)2)-ex2…

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• MATH

g(x)=∫1+2×1−2x(tsin(t))dtg(x)=∫1-2×1+2x(tsin(t))dt Find the derivative of the function.

Hello! Let’s temporarily denote the antiderivative of tsin(t)tsin(t) as A(t).A(t).Then the integral is equal to A(1+2x)−A(1−2x)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)).A′(1+2x)⋅2-A′(1-2x)⋅(-2)=2⋅(A′(1+2x)+A′(1-2x))….

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• MATH

g(x)=∫3x2x(u2−1u2+1)dug(x)=∫2x3x(u2-1u2+1)du Find the derivative of the function.

g(x)=∫3x2xu2−1u2+1dug(x)=∫2x3xu2-1u2+1du g(x)=∫02xu2−1u2+1du+∫3x0u2−1u2+1dug(x)=∫2x0u2-1u2+1du+∫03xu2-1u2+1du g(x)=−∫2x0u2−1u2+1du+∫3x0u2−1u2+1dug(x)=-∫02xu2-1u2+1du+∫03xu2-1u2+1du…

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• MATH

∫2ππ6(cos(x))dx∫π62π(cos(x))dx Evaluate the integral and interpret it as a difference of…

The integral actually represents difference between red area and green area shown in the image below i.e. ∫2ππ6cos(x)dx=P1−P2+P3∫π62πcos(x)dx=P1-P2+P3…

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• MATH

∫2−1(x3)dx∫-12(x3)dx Evaluate the integral and interpret it as a difference of areas….

∫2−1(x3)dx∫-12(x3)dx Let us first solve: ∫(x3)dx=(x3+13+1)+c=(x44)+C∫(x3)dx=(x3+13+1)+c=(x44)+C Looking at the graph below we can say that the area covered by the curve x3x3 for −1≤x≤2-1≤x≤2 is…

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• MATH

y=sec2(x),0≤x≤(π3)y=sec2(x),0≤x≤(π3) Use a graph to give a rough estimate of the area of the…

y=sec2(x)y=sec2(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=≈(410)(π3⋅4)≈(410)(π3⋅4) Area of the…

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• MATH

y=sin(x),0≤x≤πy=sin(x),0≤x≤π Use a graph to give a rough estimate of the area of the region…

Refer the graph in the attached image. From the graph it appears that the area is ≈≈ 2/3 of the rectangle. Area of the region= ≈≈ 2/3(Area of rectangle) Area of the region…

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• MATH

y=x−4,1≤x≤6y=x-4,1≤x≤6 Use a graph to give a rough estimate of the area of the region…

Refer the graph in the attached image. From the graph, Area of the region that lies beneath the curve ≈≈ 6/100(Area of Rectangle) Area of the region ≈≈ 6/100(5*1) ≈≈ 0.30 Exact Area of the…

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• MATH

y=x−−√3,0≤x≤27y=x3,0≤x≤27 Use a graph to give a rough estimate of the area of the…

Refer the graph in the attached image. From the graph it appears that area of the region is ≈≈ 70% of the rectangle. Area of the region=≈≈70/100(Area of Rectangle) Area of the region…

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• MATH

∫12√12(41−x2−−−−−√)dx∫1212(41-x2)dx Evaluate the integral.

Hello! This integral is a table one, ∫(41−x2−−−−−√)dx=4arcsin(x)+C.∫(41-x2)dx=4arcsin(x)+C. Therefore the definite integral is equal to 4⋅(arcsin(12–√)−arcsin(12))=4⋅(π4−π6)=4⋅π12=π3≈1.047.4⋅(arcsin(12)-arcsin(12))=4⋅(π4-π6)=4⋅π12=π3≈1.047.

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• MATH

∫1−1(eu+1)du∫-11(eu+1)du Evaluate the integral.

∫1−1eu+1du∫-11eu+1du To evaluate this, apply the formula ∫exdx=ex∫exdx=ex . =eu+1∣∣1−1=eu+1∣-11 Then, plug-in the limits of the integral as follows F(x)=∫baf(x)dx=F(b)−F(a)F(x)=∫abf(x)dx=F(b)-F(a) ….

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• MATH

∫21(4+u2u3)du∫12(4+u2u3)du Evaluate the integral.

You need to evaluate the definite integral using the fundamental theorem of calculus, such that: ∫baf(u)du=F(b)−F(a)∫abf(u)du=F(b)-F(a)

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• MATH

∫3√13√(81+x2)dx∫133(81+x2)dx Evaluate the integral.

You need to evaluate the definite integral using the fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a)

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• MATH

∫10(cosh(t))dt∫01(cosh(t))dt Evaluate the integral.

We have to evaluate the integral ∫10cosh(t)dt∫01cosh(t)dt  We know that the integral of cosh(t) = sinh(t) . Therefore we can write, ∫10cosh(t)dt=[sinh(t)]10∫01cosh(t)dt=[sinh(t)]01…

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