Further problems 3

Q1

Determine the antiderivative of the following functions.

  1. f(x)=xf(x)=x
  2. g(x)=x2g(x)=x^2
  3. h(x)=xh(x)=\sqrt{x}
  4. i(x)=1x3i(x)=\frac{1}{x^3}
  5. j(x)=1xj(x)=\frac{1}{\sqrt{x}}
  6. k(x)=1k(x)=1 (for all xx)
  7. l(x)=0l(x)=0 (for all xx)
  8. v(x)=3v(x)=3 (for all xx)
  9. m(x)=3x2+4x4−2x−1m(x)=3x^2+4x^4-2x-1
  10. n(x)=2x2+1n(x)=\frac{2}{x^2}+1
  11. o(x)=3x4−2xo(x)=3\sqrt[4]{x}-\frac{2}{x}
  12. p(x)=3sin⁡(x)−πexp(x)=3\sin(x)-\pi e^x
  13. q(x)=2x2⋅x5q(x)=2 x^2\cdot x^5
  14. r(x)=3x(x−1)r(x)=3x(x-1)
  15. s(x)=x2+xx2s(x)=\frac{x^2+x}{x^2}
  16. t(x)=7x(x2−1)t(x)=7\sqrt{x}(x^2-1)
  17. u(x)=(2x−3)2u(x)=(2x-3)^2
Q2

Determine the following integrals using theorem of calculus.

  1. ∫132x2 dx\int_1^3 2x^2 \, dx
  2. ∫−13(x2+2) dx\int_{-1}^3 (x^2+2) \, dx
  3. ∫010.1x dx\int_0^1 0.1 \sqrt{x} \, dx
  4. ∫12(2x+x) dx\int_{1}^2 (\frac{2}{x}+x) \, dx
  5. ∫12(x−1)2 dx\int_{1}^2 (x-1)^2 \, dx
  6. ∫015ex dx\int_0^1 5e^x \, dx
  7. ∫−π/4π/4(cos⁡(x)−sin⁡(x)) dx\int_{-\pi/4}^{\pi/4} \left( \cos(x)-\sin(x) \right) \, dx
Q3

Determine the signed area between the graph of the function f(x)=x3−x4+5f(x)=x^3-x^4+5 and the xx-axis for a=−1a=-1 and b=1b=1.

Q4

The area from a=0a=0 to b=ub=u beneath the graph of the function f(x)=xf(x)=\sqrt{x} is 1010. Find the value of uu.

Q5

A car moves along a straight road such that at any point tt in time (in sec) the car moves with the instantaneous speed v(t)=1tv(t)=\frac{1}{t} (in units m/secm/sec). What distance has the car covered in the time interval from 1s1s to 100sec100sec? Hint: Recall the meaning of the sum of bars in this context.

Q6

Consider the functions g(x)=2x⋅e(x2)g(x)=2x \cdot e^{(x^2)} and h(x)=e(x2)h(x)=e^{(x^2)}.

  1. Show that the derivative of hh is gg.
  2. Determine the integral ∫012x⋅e(x2) dx\int_0^1 2x \cdot e^{(x^2)} \,dx
Show

Solutions

A1
  1. f(x)=x=x1 (n=1)→F(x)=11+1x1+1=12x2f(x)=x=x^1 \, (n=1) \rightarrow F(x)=\frac{1}{1+1}x^{1+1}=\frac{1}{2}x^2
  2. g(x)=x2 (n=2)→G(x)=12+1x2+1=13x3g(x)=x^2 \, (n=2) \rightarrow G(x)=\frac{1}{2+1}x^{2+1}=\frac{1}{3}x^3
  3. h(x)=x=x1/2 (n=1/2)→H(x)=11/2+1x1/2+1=11.5x1.5=23x1.5h(x)=\sqrt{x}=x^{1/2} \, (n=1/2) \rightarrow H(x)=\frac{1}{1/2+1}x^{1/2+1}=\frac{1}{1.5}x^{1.5}=\frac{2}{3}x^{1.5}
  4. i(x)=1x3=x−3 (n=−3)→I(x)=1−3+1x−3+1=1−2x−2=−12x2i(x)=\frac{1}{x^3}=x^{-3} \, (n=-3) \rightarrow I(x)=\frac{1}{-3+1}x^{-3+1}=\frac{1}{-2}x^{-2}=-\frac{1}{2x^2}
  5. j(x)=1x=1x1/2=x−1/2 (n=−1/2)→J(x)=1−1/2+1x−1/2+1=11/2x1/2=2xj(x)=\frac{1}{\sqrt{x}}=\frac{1}{x^{1/2}}=x^{-1/2} \, (n=-1/2) \rightarrow J(x)=\frac{1}{-1/2+1}x^{-1/2+1}=\frac{1}{1/2}x^{1/2}=2\sqrt{x}
  6. k(x)=1=x0 (n=0)→K(x)=10+1x0+1=x1=xk(x)=1=x^0 \, (n=0) \rightarrow K(x)=\frac{1}{0+1}x^{0+1}=x^1=x
  7. l(x)=0l(x)=0. The function whose derivative equals 00 for all xx must be a constant function. So for example L(x)=1L(x)=1 for all xx.
  8. v(x)=3=3⋅x0→V(x)=3⋅10+1x0+1=3xv(x)=3=3\cdot x^0 \rightarrow V(x)=3\cdot \frac{1}{0+1}x^{0+1}=3x
  9. m(x)=3x2+4x4−2x−1=3x2+4x4−2x1−1⋅x0→M(x)=3⋅12+1x2+1+4⋅14+1x4+1−211+1x1+1−1⋅10+1x0+1=x3+45x5−x2−xm(x)=3x^2+4x^4-2x-1 = 3x^2+4x^4-2x^1-1\cdot x^0\rightarrow M(x)=3\cdot \frac{1}{2+1}x^{2+1}+4\cdot \frac{1}{4+1}x^{4+1}-2\frac{1}{1+1}x^{1+1}-1\cdot \frac{1}{0+1}x^{0+1}= x^3+\frac{4}{5}x^5-x^2-x
  10. n(x)=2x2+1=2⋅1x2+x0=2⋅x−2+x0→N(x)=2⋅1−2+1x−2+1+10+1x0+1=−2x−1+x=−2x+xn(x)=\frac{2}{x^2}+1 = 2\cdot \frac{1}{x^2}+x^0= 2\cdot x^{-2}+x^0 \rightarrow N(x) = 2\cdot\frac{1}{-2+1}x^{-2+1}+\frac{1}{0+1}x^{0+1}=-2x^{-1}+x=-\frac{2}{x}+x
  11. o(x)=3x4−2x=3x1/4−21x→O(x)=3⋅11/4+1x1/4+1−2⋅ln⁡(x)=125x5/4−2ln⁡(x)o(x)=3\sqrt[4]{x}-\frac{2}{x} =3 x^{1/4} - 2 \frac{1}{x} \rightarrow O(x) = 3\cdot \frac{1}{1/4+1}x^{1/4+1} -2\cdot \ln(x) = \frac{12}{5} x^{5/4}-2\ln(x)
  12. p(x)=3sin⁡(x)−πex→P(x)=3⋅(−cos⁡(x))−π⋅ex=−3cos⁡(x)−πexp(x)=3\sin(x)-\pi e^x \rightarrow P(x)=3\cdot (-\cos(x))-\pi \cdot e^x=-3\cos(x)-\pi e^x
  13. q(x)=2x2⋅x5=2x7→Q(x)=2⋅17+1x7+1=14x8q(x)=2 x^2\cdot x^5 = 2x^7 \rightarrow Q(x)=2\cdot \frac{1}{7+1} x^{7+1}=\frac{1}{4}x^8
  14. r(x)=3x(x−1)=3x2−3x1→R(x)=3⋅12+1x2+1−311+1x1+1=x3−32x2r(x)=3x(x-1)=3x^2-3x^1 \rightarrow R(x)=3\cdot \frac{1}{2+1}x^{2+1}-3\frac{1}{1+1}x^{1+1}=x^3-\frac{3}{2}x^2
  15. s(x)=x2+xx2=x2x2+xx2=1+1x=x0+1x→S(x)=10+1x0+1+ln⁡(x)=x+ln⁡(x)s(x)=\frac{x^2+x}{x^2} =\frac{x^2}{x^2}+\frac{x}{x^2}=1+\frac{1}{x} =x^0+\frac{1}{x}\rightarrow S(x)=\frac{1}{0+1}x^{0+1}+\ln(x)=x+\ln(x)
  16. t(x)=7x(x2−1)=7x1/2x2−7x1/2=7x5/2−7x1/2→T(x)=7⋅15/2+1x5/2+1−7⋅11/2+1x1/2+1=2x7/2−143x3/2t(x)=7\sqrt{x}(x^2-1) = 7x^{1/2}x^2-7x^{1/2}=7x^{5/2}-7x^{1/2} \rightarrow T(x)=7\cdot \frac{1}{5/2+1}x^{5/2+1}-7\cdot \frac{1}{1/2+1}x^{1/2+1} = 2x^{7/2}-\frac{14}{3}x^{3/2}
  17. u(x)=(2x−3)2=4x2−12x+9=4x2−12x1+9x0→U(x)=4⋅12+1x2+1−12⋅11+1x1+1+9⋅10+1x0+1=43x3−6x2+9xu(x)=(2x-3)^2 = 4x^2-12x+9=4x^2-12x^1+9x^0 \rightarrow U(x)=4\cdot \frac{1}{2+1}x^{2+1}-12\cdot \frac{1}{1+1}x^{1+1}+9\cdot \frac{1}{0+1}x^{0+1}=\frac{4}{3}x^3-6x^2+9x
A2
  1. Antiderivative F(x)=23x3→∫132x2 dx=F(3)−F(1)=23⋅33−23⋅13=17.3ˉ‾F(x)=\frac{2}{3}x^3\\ \rightarrow \int_1^3 2x^2 \, dx = F(3)-F(1) = \frac{2}{3}\cdot 3^3-\frac{2}{3}\cdot 1^3=\underline{17.\bar 3}
  2. F(x)=13x3+2x→∫−13(x2+2) dx=F(3)−F(−1)=(13⋅33+2⋅3)−(13⋅(−1)3+2⋅(−1))=17.3ˉ‾F(x)=\frac{1}{3}x^3+2x \\ \rightarrow \int_{-1}^3 (x^2+2) \, dx = F(3)-F(-1)=\left(\frac{1}{3}\cdot 3^3+2\cdot 3\right) - \left(\frac{1}{3}\cdot (-1)^3+2\cdot (-1)\right) = \underline{17.\bar 3}
  3. F(x)=0.06ˉ⋅x3/2→∫010.1x dx=F(1)−F(0)=(0.06ˉ⋅13/2)−(0.06ˉ⋅03/2)=0.06ˉ‾F(x)=0.0\bar 6\cdot x^{3/2} \rightarrow \int_0^1 0.1 \sqrt{x} \, dx = F(1)-F(0)=(0.0\bar 6\cdot 1^{3/2})-(0.0\bar 6\cdot 0^{3/2}) = \underline{0.0\bar 6}
  4. F(x)=2ln⁡(x)+12x2→∫12(2x+x) dx=F(2)−F(1)=(2ln⁡(2)+1222)−(2ln⁡(1)+1212)=2.886...‾F(x)= 2\ln(x)+\frac{1}{2}x^2 \rightarrow \int_{1}^2 (\frac{2}{x}+x) \, dx = F(2)-F(1)= \left( 2\ln(2)+\frac{1}{2}2^2 \right) - \left( 2\ln(1)+\frac{1}{2}1^2 \right) =\underline{2.886...}
  5. F(x)=13x3−x2+x→∫12(x−1)2 dx=F(2)−F(1)=(1323−22+2)−(1313−12+1)=13‾F(x)=\frac{1}{3}x^3-x^2+x \rightarrow \int_{1}^2 (x-1)^2 \, dx = F(2)-F(1) = \left( \frac{1}{3}2^3-2^2+2 \right) -\left( \frac{1}{3}1^3-1^2+1 \right) = \underline{\frac{1}{3}}
  6. F(x)=5ex→∫015ex dx=F(1)−F(0)=5e1−5e0=8.591...‾F(x)= 5e^x \rightarrow \int_0^1 5e^x \, dx = F(1)-F(0)=5e^1-5e^0=\underline{8.591...}
  7. F(x)=sin⁡(x)+cos⁡(x)→∫−π/4π/4(cos⁡(x)−sin⁡(x)) dx=F(π/4)−F(−π/4)=(sin⁡(π/4)+cos⁡(π/4))−(sin⁡(−π/4)+cos⁡(−π/4))=1.414...‾=2‾F(x)=\sin(x)+\cos(x) \rightarrow \int_{-\pi/4}^{\pi/4} \left(\cos(x)-\sin(x) \right) \, dx = F(\pi/4)-F(-\pi/4) = \left( \sin(\pi/4)+\cos(\pi/4) \right) - \left( \sin(-\pi/4)+\cos(-\pi/4) \right) = \underline{1.414...} = \underline{\sqrt{2}}
A3

A=∫−11(x3−x4+5) dxA=\int_{-1}^1 (x^3-x^4+5) \, dx. The antiderivative is F(x)=14x4−15x5+5xF(x)=\frac{1}{4}x^4-\frac{1}{5}x^5+5x. Thus, we have

A=∫−11(x3−x4+5) dx=F(1)−F(−1)=(14−15+5)−(14+15−5)=9.6‾\begin{array}{lll} A &=&\int_{-1}^1 (x^3-x^4+5) \, dx \\ &=& F(1)-F(-1)\\ &=& (\frac{1}{4}-\frac{1}{5}+5)-(\frac{1}{4}+\frac{1}{5}-5)\\ &=&\underline{9.6} \end{array}
A4

The area is A=∫0ux dxA=\int_0^u \sqrt{x}\, dx. The antiderivative of f(x)=x=x1/2f(x)=\sqrt{x}=x^{1/2} is F(x)=11/2+1x1/2+1=23x1.5F(x)=\frac{1}{1/2+1} x^{1/2+1}=\frac{2}{3}x^{1.5}. Thus, find uu with

A=∫0ux dx=F(u)−F(0)=23u1.5−0=10A=\int_0^u \sqrt{x}\, dx =F(u)-F(0)=\frac{2}{3}u^{1.5}-0=10

So we have to solve the equation

23u1.5=10∣⋅3/2u1.5=15∣(.)1/1.5u=151/1.5=6.08...‾\begin{array}{lll} \frac{2}{3}u^{1.5} & = & 10 \quad | \cdot 3/2\\ u^{1.5} & = & 15 \quad | (.)^{1/1.5} \\ u &=& 15^{1/1.5}\\ &= & \underline{6.08...} \end{array}
A5

The distance is given by the area under the curve, that is, the integral

s=∫11001t dt=ln⁡(100)−ln⁡(1)=4.605m‾s=\int_1^{100} \frac{1}{t} \, dt = \ln(100)-\ln(1)=\underline{4.605m}
A6

  1. Using the chain rule with u(x)=exu(x)=e^x and i(x)=x2i(x)=x^2, we get

    h′(x)=e(x2)⋅2xh'(x)=e^{(x^2)}\cdot 2x

    which is gg.

  2. As h′=gh'=g, it follows that hh is an antiderivative of gg, and thus

    ∫012x⋅e(x2) dx=h(1)−h(0)=e1−e0=e−1‾\int_0^1 2x\cdot e^{(x^2)} \,dx = h(1)-h(0)=e^1-e^0=\underline{e-1}