Logo

AskSia

Plus

1. Evaluate the integral. Z a) (1 − 2x)9 dx Z b) sec2 θ tan3 θ dθ Z c) x(2...
May 15, 2024
1. Evaluate the integral. Z a) (1 − 2x)9 dx Z b) sec2 θ tan3 θ dθ Z c) x(2x + 5)8 dx Z 1 0 3 √ d) 1 + 7x dx Z π/4 −π/4 􀀀 x3 + x4 tan x   e) dx Z a 0 x √ f) a2 − x2 dx 2. If f is continuous and Z 4 0 f(x) dx = 10, find Z 2 0 f(2x) dx. 3. Evaluate the integral. (a). Z (x − 1) sin πx dx (b). Z z 10z dz (c). Z 1 0 (x2 + 1)e−x dx 4. A particle that moves along a straight line has velocity v(t) = t2e−t meters per second after t seconds. How far will it travel during the first t seconds? 5. Evaluate the integral. (a). Z sin3 θ cos4 θ dθ (b). Z 2π 0 sin2( 1 3 x) dx (c). Z sin2(1/t) t2 dt (d). Z x sin3 x dx (e). Z π/4 0 sec6 x tan6 x dx (f). Z tan2 x sec x dx (g). Z π/2 0 cos 5t cos 10t dt
Generated Graph
Solution by Steps
step 1
To integrate (1−2x)9 dx(1 - 2x)^9 \, dx, we use the substitution u=1−2xu = 1 - 2x, hence du=−2 dxdu = -2 \, dx
step 2
The integral becomes ∫(1−2x)9 dx=−12∫u9 du\int (1 - 2x)^9 \, dx = -\frac{1}{2} \int u^9 \, du
step 3
Integrating u9u^9 gives −12⋅u1010+C=−120(1−2x)10+C-\frac{1}{2} \cdot \frac{u^{10}}{10} + C = -\frac{1}{20} (1 - 2x)^{10} + C
# Part (b)
step 1
To integrate sec⁡2(θ)tan⁡3(θ) dθ\sec^2(\theta) \tan^3(\theta) \, d\theta, we use the substitution u=tan⁡(θ)u = \tan(\theta), hence du=sec⁡2(θ) dθdu = \sec^2(\theta) \, d\theta
step 2
The integral becomes ∫u3 du=u44+C=tan⁡4(θ)4+C\int u^3 \, du = \frac{u^4}{4} + C = \frac{\tan^4(\theta)}{4} + C
# Part (c)
step 1
To integrate x(2x+5)8 dxx(2x + 5)^8 \, dx, we use the substitution u=2x+5u = 2x + 5, hence du=2 dxdu = 2 \, dx and x=u−52x = \frac{u - 5}{2}
step 2
The integral becomes ∫u−52u8⋅du2=14∫(u9−5u8) du\int \frac{u - 5}{2} u^8 \cdot \frac{du}{2} = \frac{1}{4} \int (u^9 - 5u^8) \, du
step 3
Integrating gives 14(u1010−5u99)+C=(2x+5)1040−5(2x+5)936+C\frac{1}{4} \left( \frac{u^{10}}{10} - 5 \frac{u^9}{9} \right) + C = \frac{(2x + 5)^{10}}{40} - \frac{5(2x + 5)^9}{36} + C
# Part (d)
step 1
To integrate ∫01(3+7x)1/3 dx\int_0^1 (3 + 7x)^{1/3} \, dx, we use the substitution u=3+7xu = 3 + 7x, hence du=7 dxdu = 7 \, dx
step 2
The integral becomes ∫310u1/3⋅du7=17∫u1/3 du\int_3^{10} u^{1/3} \cdot \frac{du}{7} = \frac{1}{7} \int u^{1/3} \, du
step 3
Integrating u1/3u^{1/3} gives 17⋅34u4/3∣310=328(104/3−34/3)≈1.8447\frac{1}{7} \cdot \frac{3}{4} u^{4/3} \bigg|_3^{10} = \frac{3}{28} \left( 10^{4/3} - 3^{4/3} \right) \approx 1.8447
# Part (e)
step 1
To integrate ∫−π/4π/4(x3+x4tan⁡(x)) dx\int_{-\pi/4}^{\pi/4} (x^3 + x^4 \tan(x)) \, dx, we note that x3x^3 is an odd function and x4tan⁡(x)x^4 \tan(x) is an odd function over the symmetric interval [−π/4,π/4][-\pi/4, \pi/4]
step 2
The integral of an odd function over a symmetric interval is zero, hence ∫−π/4π/4(x3+x4tan⁡(x)) dx=0\int_{-\pi/4}^{\pi/4} (x^3 + x^4 \tan(x)) \, dx = 0
# Part (f)
step 1
To integrate ∫0a(a2−x2)1/2 dx\int_0^a (a^2 - x^2)^{1/2} \, dx, we use the substitution x=asin⁡(θ)x = a \sin(\theta), hence dx=acos⁡(θ) dθdx = a \cos(\theta) \, d\theta
step 2
The integral becomes ∫0π/2a2cos⁡2(θ) dθ=a2∫0π/2cos⁡2(θ) dθ\int_0^{\pi/2} a^2 \cos^2(\theta) \, d\theta = a^2 \int_0^{\pi/2} \cos^2(\theta) \, d\theta
step 3
Using the identity cos⁡2(θ)=1+cos⁡(2θ)2\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}, the integral becomes a2∫0π/21+cos⁡(2θ)2 dθ=a22[θ+sin⁡(2θ)2]0π/2=πa24a^2 \int_0^{\pi/2} \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{a^2}{2} \left[ \theta + \frac{\sin(2\theta)}{2} \right]_0^{\pi/2} = \frac{\pi a^2}{4}
Question 1 Answer
A
Question 2
step 1
Given ∫04f(x) dx=10\int_0^4 f(x) \, dx = 10, we need to find ∫02f(2x) dx\int_0^2 f(2x) \, dx
step 2
Using the substitution u=2xu = 2x, hence du=2 dxdu = 2 \, dx, the integral becomes ∫02f(2x) dx=12∫04f(u) du=12⋅10=5\int_0^2 f(2x) \, dx = \frac{1}{2} \int_0^4 f(u) \, du = \frac{1}{2} \cdot 10 = 5
Question 2 Answer
B
Question 3 # Part (a)
step 1
To integrate (x−1)sin⁡(πx) dx(x - 1) \sin(\pi x) \, dx, we use integration by parts with u=x−1u = x - 1 and dv=sin⁡(πx) dxdv = \sin(\pi x) \, dx
step 2
Then du=dxdu = dx and v=−cos⁡(πx)πv = -\frac{\cos(\pi x)}{\pi}
step 3
The integral becomes −(x−1)cos⁡(πx)π+∫cos⁡(πx)π dx-\frac{(x - 1) \cos(\pi x)}{\pi} + \int \frac{\cos(\pi x)}{\pi} \, dx
step 4
Integrating cos⁡(πx)π\frac{\cos(\pi x)}{\pi} gives −sin⁡(πx)π2-\frac{\sin(\pi x)}{\pi^2}, hence the integral is −(x−1)cos⁡(πx)π−sin⁡(πx)π2+C-\frac{(x - 1) \cos(\pi x)}{\pi} - \frac{\sin(\pi x)}{\pi^2} + C
# Part (b)
step 1
To integrate z⋅10z dzz \cdot 10z \, dz, we rewrite it as 10z2 dz10z^2 \, dz
step 2
Integrating 10z210z^2 gives 10z33+C\frac{10z^3}{3} + C
# Part (c)
step 1
To integrate ∫01(x2+1)e−x dx\int_0^1 (x^2 + 1) e^{-x} \, dx, we use integration by parts with u=x2+1u = x^2 + 1 and dv=e−x dxdv = e^{-x} \, dx
step 2
Then du=2x dxdu = 2x \, dx and v=−e−xv = -e^{-x}
step 3
The integral becomes −(x2+1)e−x∣01+∫012xe−x dx-(x^2 + 1) e^{-x} \bigg|_0^1 + \int_0^1 2x e^{-x} \, dx
step 4
Evaluating the first part gives −(12+1)e−1+(02+1)e0=−2/e+1-(1^2 + 1) e^{-1} + (0^2 + 1) e^0 = -2/e + 1
step 5
For the second part, we use integration by parts again with u=2xu = 2x and dv=e−x dxdv = e^{-x} \, dx
step 6
Then du=2 dxdu = 2 \, dx and v=−e−xv = -e^{-x}
step 7
The integral becomes −2xe−x∣01+∫012e−x dx=−2/e+2(1−1/e)=2−4/e-2x e^{-x} \bigg|_0^1 + \int_0^1 2 e^{-x} \, dx = -2/e + 2(1 - 1/e) = 2 - 4/e
step 8
Combining both parts, the integral is 1−2/e+2−4/e=3−6/e≈0.792721 - 2/e + 2 - 4/e = 3 - 6/e \approx 0.79272
Question 3 Answer
C
Question 4
step 1
Given the velocity v(t)=t2e−tv(t) = t^2 e^{-t}, we need to find the distance traveled during the first tt seconds, which is ∫0tt2e−t dt\int_0^t t^2 e^{-t} \, dt
step 2
Using integration by parts with u=t2u = t^2 and dv=e−t dtdv = e^{-t} \, dt, we get du=2t dtdu = 2t \, dt and v=−e−tv = -e^{-t}
step 3
The integral becomes −t2e−t∣0t+∫0t2te−t dt-t^2 e^{-t} \bigg|_0^t + \int_0^t 2t e^{-t} \, dt
step 4
Evaluating the first part gives −t2e−t+0=−t2e−t-t^2 e^{-t} + 0 = -t^2 e^{-t}
step 5
For the second part, we use integration by parts again with u=2tu = 2t and dv=e−t dtdv = e^{-t} \, dt
step 6
Then du=2 dtdu = 2 \, dt and v=−e−tv = -e^{-t}
step 7
The integral becomes −2te−t∣0t+∫0t2e−t dt=−2te−t+2(1−e−t)=2−2e−t−2te−t-2t e^{-t} \bigg|_0^t + \int_0^t 2 e^{-t} \, dt = -2t e^{-t} + 2(1 - e^{-t}) = 2 - 2e^{-t} - 2t e^{-t}
step 8
Combining both parts, the distance is 2−e−t(t2+2t+2)2 - e^{-t} (t^2 + 2t + 2)
Question 4 Answer
D
Question 5 # Part (a)
step 1
To integrate sin⁡3(θ)cos⁡4(θ) dθ\sin^3(\theta) \cos^4(\theta) \, d\theta, we use the identity sin⁡3(θ)=sin⁡(θ)(1−cos⁡2(θ))cos⁡4(θ)\sin^3(\theta) = \sin(\theta) (1 - \cos^2(\theta)) \cos^4(\theta)
step 2
The integral becomes ∫sin⁡(θ)cos⁡4(θ) dθ−∫sin⁡(θ)cos⁡6(θ) dθ\int \sin(\theta) \cos^4(\theta) \, d\theta - \int \sin(\theta) \cos^6(\theta) \, d\theta
step 3
Using the substitution u=cos⁡(θ)u = \cos(\theta), hence du=−sin⁡(θ) dθdu = -\sin(\theta) \, d\theta, the integrals become −∫u4 du+∫u6 du-\int u^4 \, du + \int u^6 \, du
step 4
Integrating gives −u55+u77+C=−cos⁡5(θ)5+cos⁡7(θ)7+C-\frac{u^5}{5} + \frac{u^7}{7} + C = -\frac{\cos^5(\theta)}{5} + \frac{\cos^7(\theta)}{7} + C
# Part (b)
step 1
To integrate ∫02πsin⁡2(x3) dx\int_0^{2\pi} \sin^2(\frac{x}{3}) \, dx, we use the identity sin⁡2(x3)=1−cos⁡(2x3)2\sin^2(\frac{x}{3}) = \frac{1 - \cos(\frac{2x}{3})}{2}
step 2
The integral becomes ∫02π1−cos⁡(2x3)2 dx=12∫02π1 dx−12∫02πcos⁡(2x3) dx\int_0^{2\pi} \frac{1 - \cos(\frac{2x}{3})}{2} \, dx = \frac{1}{2} \int_0^{2\pi} 1 \, dx - \frac{1}{2} \int_0^{2\pi} \cos(\frac{2x}{3}) \, dx
step 3
The first integral is 12⋅2π=π\frac{1}{2} \cdot 2\pi = \pi
step 4
The second integral is zero because cos⁡(2x3)\cos(\frac{2x}{3}) is periodic with period 3π3\pi
step 5
Hence, the integral is π\pi
# Part (c)
step 1
To integrate ∫sin⁡2(1t)t2 dt\int \frac{\sin^2(\frac{1}{t})}{t^2} \, dt, we use the substitution u=1tu = \frac{1}{t}, hence du=−1t2 dtdu = -\frac{1}{t^2} \, dt
step 2
The integral becomes −∫sin⁡2(u) du-\int \sin^2(u) \, du
step 3
Using the identity sin⁡2(u)=1−cos⁡(2u)2\sin^2(u) = \frac{1 - \cos(2u)}{2}, the integral becomes −∫1−cos⁡(2u)2 du=−12∫1 du+12∫cos⁡(2u) du-\int \frac{1 - \cos(2u)}{2} \, du = -\frac{1}{2} \int 1 \, du + \frac{1}{2} \int \cos(2u) \, du
step 4
Integrating gives −u2+sin⁡(2u)4+C=−12t+sin⁡(2t)4+C-\frac{u}{2} + \frac{\sin(2u)}{4} + C = -\frac{1}{2t} + \frac{\sin(\frac{2}{t})}{4} + C
# Part (d)
step 1
To integrate xsin⁡3(x) dxx \sin^3(x) \, dx, we use the identity sin⁡3(x)=sin⁡(x)(1−cos⁡2(x))\sin^3(x) = \sin(x) (1 - \cos^2(x))
step 2
The integral becomes ∫xsin⁡(x) dx−∫xsin⁡(x)cos⁡2(x) dx\int x \sin(x) \, dx - \int x \sin(x) \cos^2(x) \, dx
step 3
Using integration by parts for the first integral with u=xu = x and dv=sin⁡(x) dxdv = \sin(x) \, dx, we get du=dxdu = dx and v=−cos⁡(x)v = -\cos(x)
step 4
The integral becomes −xcos⁡(x)+∫cos⁡(x) dx=−xcos⁡(x)+sin⁡(x)-x \cos(x) + \int \cos(x) \, dx = -x \cos(x) + \sin(x)
step 5
For the second integral, we use the substitution u=cos⁡(x)u = \cos(x), hence du=−sin⁡(x) dxdu = -\sin(x) \, dx
step 6
The integral becomes −∫xu2 du-\int x u^2 \, du
step 7
Using integration by parts with u=xu = x and dv=u2 dudv = u^2 \, du, we get du=dxdu = dx and v=u33v = \frac{u^3}{3}
step 8
The integral becomes −xu33+∫u33 dx=−xcos⁡3(x)3+cos⁡3(x)9-\frac{x u^3}{3} + \int \frac{u^3}{3} \, dx = -\frac{x \cos^3(x)}{3} + \frac{\cos^3(x)}{9}
step 9
Combining both parts, the integral is −xcos⁡3(x)3+cos⁡3(x)9−xcos⁡(x)+sin⁡(x)+C-\frac{x \cos^3(x)}{3} + \frac{\cos^3(x)}{9} - x \cos(x) + \sin(x) + C
# Part (e)
step 1
To integrate ∫0π/4sec⁡6(x)tan⁡6(x) dx\int_0^{\pi/4} \sec^6(x) \tan^6(x) \, dx, we use the substitution u=tan⁡(x)u = \tan(x), hence du=sec⁡2(x) dxdu = \sec^2(x) \, dx
step 2
The integral becomes ∫01u6sec⁡4(x) du=∫01u6(1+u2)2 du\int_0^1 u^6 \sec^4(x) \, du = \int_0^1 u^6 (1 + u^2)^2 \, du
step 3
Expanding (1+u2)2(1 + u^2)^2 gives 1+2u2+u41 + 2u^2 + u^4, hence the integral becomes ∫01(u6+2u8+u10) du\int_0^1 (u^6 + 2u^8 + u^{10}) \, du
step 4
Integrating gives u77+2u99+u1111∣01=17+29+111=316693≈0.45599\frac{u^7}{7} + \frac{2u^9}{9} + \frac{u^{11}}{11} \bigg|_0^1 = \frac{1}{7} + \frac{2}{9} + \frac{1}{11} = \frac{316}{693} \approx 0.45599
# Part (f)
step 1
To integrate tan⁡2(x)sec⁡(x) dx\tan^2(x) \sec(x) \, dx, we use the identity tan⁡2(x)=sec⁡2(x)−1\tan^2(x) = \sec^2(x) - 1
step 2
The integral becomes ∫(sec⁡3(x)−sec⁡(x)) dx\int (\sec^3(x) - \sec(x)) \, dx
step 3
Using the substitution u=sec⁡(x)+tan⁡(x)u = \sec(x) + \tan(x), hence du=(sec⁡(x)tan⁡(x)+sec⁡2(x)) dxdu = (\sec(x) \tan(x) + \sec^2(x)) \, dx
step 4
The integral becomes ∫u du=u22+C=(sec⁡(x)+tan⁡(x))22+C\int u \, du = \frac{u^2}{2} + C = \frac{(\sec(x) + \tan(x))^2}{2} + C
# Part (g)
step 1
To integrate ∫0π/2cos⁡(5t)cos⁡(10t) dt\int_0^{\pi/2} \cos(5t) \cos(10t) \, dt, we use the product-to-sum identities
© 2023 AskSia.AI all rights reserved