🚩 إبلاغ
Find the general antiderivative
أ
tanx + secx + c
ب
secx – tanx + c
ج
tanx - secx + c
د
2secx + c
تفسير الإجابة
The antiderivative of 2secx tanx is secx + c.
🚩 إبلاغ
Find the general antiderivative
أ
\(4cos⁻¹x + c\)
ب
\(4sin⁻¹x + c\)
ج
\(4tan⁻¹x + c\)
د
\(\sin⁻¹x + c\)
تفسير الإجابة
The integral of $\frac{4}{\sqrt{1-x^2}}$ dx is 4sin-1 x + c .
🚩 إبلاغ
Find the general antiderivative
أ
\(\ln|x| - 3cosx + c\)
ب
\(2ln|x| - 3sinx + c\)
ج
\(2ln|x| - 3cosx + c\)
د
\(3sinx - 2ln|x| + c\)
تفسير الإجابة
The antiderivative of $\frac{2}{x} - 3\sin x$ is $2\ln|x| - 3\cos x + c$.
🚩 إبلاغ
Determine m if $\int \frac{x^2}{1+x^m} dx = \frac{1}{3} \tan^{-1}x^3 + c$ where $m \neq 0$
أ
m = 2
ب
m = 4
ج
m = 6
د
m = 8
تفسير الإجابة
For the integral to be $\frac{1}{3} \tan^{-1}x^3 + c$, the denominator should be 1+x6 . Therefore, m=6 .
🚩 إبلاغ
Find the function f(x) satisfying the given conditions.
أ
4sinx + 4
ب
4cosx + 3
ج
4sinx + 3
د
4sinx - 3
تفسير الإجابة
Given $f'(x) = 4\cos x$, integrating gives f(x) = 4sin x + c . Since f(0) = 3 , we have 4sin(0) + c = 3 , so c=3 . Thus, f(x) = 4sin x + 3 .
🚩 إبلاغ
Find the function f(x) satisfying the given conditions. $f''(x) = 12x^2 + 2e^x$, $f'(0) = 2$, f(0) = 3
أ
x⁴ + 2ex + 3
ب
x⁴ + 2ex + 1
ج
x⁴ + 2ex – 1
د
x⁴ + 4ex + 1
تفسير الإجابة
Integrating $f''(x) = 12x^2 + 2e^x$ gives $f'(x) = 4x^3 + 2e^x + c_1$. Using $f'(0) = 2$, we get 4(0)3 + 2e0 + c1 = 2 , so 2 + c1 = 2 , which means c1 = 0 . Thus $f'(x) = 4x^3 + 2e^x$. Integrating again gives f(x) = x4 + 2e^x + c2 . Using f(0) = 3 , we get (0)4 + 2e0 + c2 = 3 , so 2 + c2 = 3 , which means c2 = 1 . Thus f(x) = x4 + 2e^x + 1 . It seems there is a typo in the provided options or question, assuming $f''(x) = 12x^2 + 2e^x$ and $f'(0)=2$, f(0)=3 , the correct answer should be x4 +2e^x+1 . However, if $f'(0)$ was 0, then f(x) = x4 +2e^x+1 . If $f'(0)$ was 4, then f(x) = x4 +2e^x-1 . Based on the options, it is likely that $f'(0)=4$ was intended or there's a typo in the $f''(x)$ expression.
🚩 إبلاغ
Use summation rules to compute the sum.
أ
2925
ب
- 875
ج
- 21980
د
7385
تفسير الإجابة
\(\text{The sum is } \sum_{i=1}^{40}(4-i^2)=\sum_{i=1}^{40}4-\sum_{i=1}^{40}i^2. \text{So } 4\times40-\frac{40(40+1)(2\times40+1)}{6}=160-22140=-21980.\)
🚩 إبلاغ
The following figure represents f(x). Find lim as n→∞ of Σ f(x_i) Δx_i in the interval [-3, 3].
🚩 إبلاغ
\(\text{Using the limit of Riemann sums for } \int_{0}^{3}(x^2+1)\,dx.\)
أ
\(\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\left(\frac{9i^2}{n^2}+1\right)\)
ب
\(\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\left(\frac{18i^2}{n^2}+1\right)\)
ج
\(\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\left(\frac{9i^2}{n}+1\right)\)
د
\(\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\frac{9i^2}{n^2}\)
🚩 إبلاغ
\(\text{Express the limit as an integral: } \lim_{n\to\infty}\frac{1}{n}\left[\sin\left(\frac{\pi}{n}\right)+\sin\left(\frac{2\pi}{n}\right)+\cdots+\sin\left(\frac{n\pi}{n}\right)\right].\)
أ
\(\int_{0}^{2}\sin(\pi x)\,dx\)
ب
\(\int_{0}^{1}\sin(\pi x)\,dx\)
ج
\(\int_{0}^{1}\sin(x)\,dx\)
د
\(\int_{0}^{1}\sin(2x)\,dx\)
🚩 إبلاغ
\(\text{Express the limit as an integral: } \lim_{n\to\infty}\frac{5}{n}\sum_{i=1}^{n}\left(2+\frac{5i}{n}\right)^2.\)
أ
\int2 ^{7}x2 \,dx
ب
\int0 ^{5}(2+x2 )\,dx
ج
\int2 ^{7}(2+x)2 \,dx
د
\int2 ^{7}(2+x2 )\,dx
🚩 إبلاغ
\(\text{Express the limit as an integral: } \lim_{n\to\infty}\frac{1}{n}\left[\ln\left(4+\frac{1}{n}\right)+\ln\left(4+\frac{2}{n}\right)+\cdots+\ln\left(4+\frac{n}{n}\right)\right].\)
أ
\(\int_{0}^{1}\ln(4+x)\,dx\)
ب
\(\int_{4}^{5}\ln(4+x)\,dx\)
ج
\(\int_{0}^{1}\ln(x)\,dx\)
د
\(\int_{0}^{1}\left(4+\ln(x)\right)\,dx\)
🚩 إبلاغ
\(\text{Assume that } \int_{1}^{3}f(x)\,dx=3 \text{ and } \int_{1}^{3}g(x)\,dx=-2.\; \text{Find } \int_{1}^{3}\left[2f(x)-g(x)\right]\,dx.\)
🚩 إبلاغ
\(\text{Assume that } \int_{1}^{3}2f(x)\,dx=6 \text{ and } \int_{1}^{3}g(x)\,dx=-2.\; \text{Find } \int_{1}^{3}\left[f(x)+4g(x)-2\right]\,dx.\)
🚩 إبلاغ
Use a geometric formula to compute the integral ∫ from 1 to 4 f(x) dx, using the graph.
🚩 إبلاغ
\(\text{Use a geometric formula to compute } \int_{0}^{2}\sqrt{4-x^2}\,dx.\)
أ
\(2\pi\)
ب
\(3\pi\)
ج
\(\pi\)
د
4
🚩 إبلاغ
\(\text{Use a geometric formula to find } a \text{ if } \int_{0}^{a}\sqrt{a^2-x^2}\,dx=4\pi.\)
🚩 إبلاغ
Find a value of c that satisfies the conclusion of the Integral Mean Value Theorem for f(x)=3x2 on [0,2].
أ
c = 2/√3
ب
c = -2/√3
ج
c = 2/√3, -2/√3
د
c = 3/√2
🚩 إبلاغ
\(\text{Compute the definite integral exactly: } \int_{1}^{4}\left(x\sqrt{x}+\frac{3}{x}\right)\,dx.\)
أ
\(\frac{62}{5}-3\ln 4\)
ب
\(\frac{62}{5}-6\ln 2\)
ج
\(\frac{62}{5}+6\ln 2\)
د
\(\frac{62}{2}+3\ln 4\)
🚩 إبلاغ
\(\text{Compute the definite integral exactly: } \int_{1}^{2}\left(4x-\frac{2}{x^2}\right)\,dx.\)
🚩 إبلاغ
\(\text{If } \int_{0}^{k}(6x+7)\,dx=48,\; \text{find } k.\)
🚩 إبلاغ
\(\text{Compute the definite integral exactly: } \int_{0}^{1/2}\frac{3}{\sqrt{1-x^2}}\,dx.\)
أ
\(\frac{\pi}{3}\)
ب
\(\frac{\pi}{2}\)
ج
\(\frac{\pi}{4}\)
د
\(\frac{\pi}{6}\)
🚩 إبلاغ
\(\text{Compute the definite integral exactly: } \int_{-1}^{1}\frac{4}{1+x^2}\,dx.\)
أ
\(\pi\)
ب
\(\frac{\pi}{2}\)
ج
\(2\pi\)
د
\(\frac{\pi}{6}\)
🚩 إبلاغ
\(\text{Find } f'(x) \text{ if } f(x)=\int_{0}^{x^2}\left(e^{-t^2}+1\right)\,dt.\)
أ
\(f'(x)=2x\left(e^{-x^2}+1\right)\)
ب
\(f'(x)=2x\left(e^{x^4}+1\right)\)
ج
\(f'(x)=2\left(e^{-x^4}+1\right)\)
د
\(f'(x)=2x\left(e^{-x^4}+1\right)\)
🚩 إبلاغ
\(\text{Find } f'(x) \text{ if } f(x)=\int_{x}^{2}\sec(t)\,dt.\)
أ
\(f'(x)=-2\sec x\)
ب
\(f'(x)=\sec x\)
ج
\(f'(x)=-\sec x\)
د
\(f'(x)=-\sec x+C\)
🚩 إبلاغ
\(\text{Find the tangent line to } y=\int_{2}^{x}\cos(\pi t^3)\,dt \text{ at } x=2.\)
أ
y=x-2
ب
y=x-1
ج
y=x+2
د
y=x
🚩 إبلاغ
\(\text{Find the tangent line to } y=\int_{0}^{x}e^{-t^2+1}\,dt \text{ at } x=0.\)
أ
y=x-e
ب
y=x+e
ج
y=ex+2
د
y=ex
🚩 إبلاغ
\(\text{Find the tangent line to } y=\int_{1}^{x^2}\sqrt{t^2+1}\,dt \text{ at } x=1.\)
أ
\(y=\sqrt{2}(x-1)\)
ب
\(y=2\sqrt{2}x-1\)
ج
\(y=\sqrt{2}x+2\)
د
\(y=2\sqrt{2}(x-1)\)
🚩 إبلاغ
\(\text{Evaluate } \int x^3\sqrt{x^4+3}\,dx.\)
أ
\(\frac{1}{15}(1+10x)^{3/2}+C\)
ب
\(\frac{1}{6}(x^4+3)^{1/2}+C\)
ج
\(\frac{1}{6}(x^4+3)^{3/2}+C\)
د
\(\frac{1}{15}(1+10x)^{1/2}+C\)
🚩 إبلاغ
\(\text{Evaluate } \int \frac{\sin x}{\sqrt{\cos x}}\,dx.\)
أ
\(-2\sqrt{\sin x}+C\)
ب
\(-2\sqrt{\cos x}+C\)
ج
-2cos x+C
د
\(\sqrt{\cos x}+C\)
🚩 إبلاغ
\(\text{Evaluate } \int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx.\)
أ
\(\sqrt{x}e^{\sqrt{x}}+C\)
ب
\(2e^{\sqrt{x}}+C\)
ج
\(2e^{2\sqrt{x}}+C\)
د
\(\frac{1}{2}e^{\sqrt{x}}+C\)
🚩 إبلاغ
\(\text{Evaluate } \int \frac{\cos(1/x)}{x^2}\,dx.\)
أ
-sin(1/x)+C
ب
cos(1/x)+C
ج
sin(1/x)+C
د
-sin(1/x2 )+C
🚩 إبلاغ
\(\text{Evaluate } \int \frac{x^2}{1+x^6}\,dx.\)
أ
\(\frac{1}{5}\ln|1+x^6|+C\)
ب
3tan-1 (x3 )+C
ج
\(\frac{1}{6}\ln|1+x^6|+C\)
د
\(\frac{1}{3}\tan^{-1}(x^3)+C\)
🚩 إبلاغ
\(\text{Evaluate } \int \frac{3\sqrt{x}}{1+x^3}\,dx.\)
أ
\(\frac{2}{5}\tan^{-1}(x^{5/2})+C\)
ب
\(\frac{2}{5}\tan^{-1}(x^{2/5})+C\)
ج
2tan-1 (x2/3 )+C
د
2tan-1 (x3/2 )+C
🚩 إبلاغ
Sketch and find the area of the region determined by y=x2 -1 and y=7-x2 .
أ
64/5
ب
64/7
ج
64/3
د
64/9
🚩 إبلاغ
Sketch and find the area of the region determined by y=√x and y=x
2 .
أ
4√2/3
ب
27/4
ج
1/3
د
4√2/9
🚩 إبلاغ
Find the area of the region bounded by x=y, x=-y, and x=1.
🚩 إبلاغ
Find the area of the region bounded by x=3y and x=2+y
2 .
أ
\int1 ^{2}(3y-2-y2 )\,dy
ب
\int0 ^{1}(3y-2-y2 )\,dy
ج
\int0 ^{2}(6-2y)\,dy
د
\(\int_{0}^{2}(\sqrt{x}-2x)\,dx\)
🚩 إبلاغ
Find the area of the region bounded by y=2x (x>0), y=3-x
2 , and x=0.
🚩 إبلاغ
Let R be bounded by y=4-2x, the x-axis, and the y-axis. Compute the volume formed by revolving R about y=4.
أ
512π/15
ب
128π/3
ج
64π/3
د
32π/3
🚩 إبلاغ
Let R be bounded by y=4-2x, the x-axis, and the y-axis. Compute the volume formed by revolving R about x=2.
أ
512π/15
ب
128π/3
ج
64π/3
د
32π/3
🚩 إبلاغ
Let R be bounded by y=x
2 and y=4. Compute the volume formed by revolving R about y=6.
أ
512π/15
ب
384π/5
ج
1408π/15
د
8π
🚩 إبلاغ
Let R be bounded by y=x
2 and y=4. Compute the volume formed by revolving R about x=-4.
أ
256π/15
ب
256π/3
ج
128π/3
د
128π/5
🚩 إبلاغ
Compute the volume of the solid formed by revolving the region bounded by y=2-x, y=0, and x=0 about the x-axis.
أ
32π/5
ب
8π/3
ج
16π/3
د
32π/3
🚩 إبلاغ
Let R be bounded by y=x
2 and y=4-x
2 . Compute the volume formed by revolving R about the x-axis.
أ
32π/5
ب
8π/3
ج
64√2π/3
د
64π/3
🚩 إبلاغ
Compute the volume of the solid formed by revolving the region bounded by y=√x, y=2, and x=0 about the y-axis.
أ
32π/5
ب
8π/3
ج
16π/3
د
32π/3
🚩 إبلاغ
Let R be bounded by y=x
2 and x=y
2 . Compute the volume formed by revolving R about x=1.
أ
3π/5
ب
4π/3
ج
11π/30
د
3π/10
🚩 إبلاغ
\(\text{Compute the arc length exactly for } y=2x-x^2,\;0\le x\le2.\)
أ
\(s=\int_{0}^{2}\sqrt{1+4(1-x)^4}\,dx\)
ب
\(s=\int_{0}^{2}\sqrt{1+2(1-x)^2}\,dx\)
ج
\(s=\int_{0}^{2}\sqrt{1+4(1-x)^2}\,dx\)
د
\(s=\int_{0}^{2}\sqrt{1+(2-2x)}\,dx\)
🚩 إبلاغ
\(\text{Compute the arc length exactly for } y=\tan x,\;0\le x\le\frac{\pi}{4}.\)
أ
\(s=\int_{0}^{\pi/4}\sec^2 x\,dx\)
ب
\(s=\int_{0}^{\pi/4}\sqrt{1+\sec^4 x}\,dx\)
ج
\(s=\int_{0}^{\pi/4}\sqrt{1+4\sec^4 x}\,dx\)
د
\(s=\int_{0}^{\pi/2}\sqrt{1+\sec^4 x}\,dx\)
🚩 إبلاغ
Compute the arc length exactly for y=lnx, 1≤x≤3.
أ
s=3.3020
ب
s=2.5020
ج
s=2.3020
د
s=2.3320
🚩 إبلاغ
\(\text{Find the surface area integral generated by revolving } y=e^x,\;0\le x\le1,\;\text{ about the x-axis}.\)
أ
\(S=2\pi\int_{0}^{1}e^x\sqrt{1+e^{2x}}\,dx\)
ب
\(S=\pi\int_{0}^{1}e^x\sqrt{1+e^{2x}}\,dx\)
ج
\(S=2\pi\int_{0}^{1}e^x\sqrt{1+e^x}\,dx\)
د
\(S=2\pi\int_{0}^{1}e^{2x}\sqrt{1+e^{2x}}\,dx\)
🚩 إبلاغ
\(\text{Find the surface area integral generated by revolving } y=\cos x,\;0\le x\le\frac{\pi}{2},\;\text{ about the x-axis}.\)
أ
\(S=2\pi\int_{0}^{\pi}\cos x\sqrt{1+\sin^2 x}\,dx\)
ب
\(S=2\pi\int_{0}^{\pi/2}\sin x\sqrt{1+\cos^2 x}\,dx\)
ج
\(S=2\pi\int_{0}^{\pi/2}\cos x\sqrt{1+\sin^2 x}\,dx\)
د
\(S=\pi\int_{0}^{\pi/2}\cos x\sqrt{1+\sin^2 x}\,dx\)
🚩 إبلاغ
A diver drops from 30 ft above the water. What is the diver's velocity at impact? Ignore air resistance.
أ
\(\text{impact velocity}=-8\sqrt{30}\;\text{ft/s}\)
ب
\(\text{impact velocity}=-10\sqrt{30}\;\text{ft/s}\)
ج
\(\text{impact velocity}=-8\sqrt{3}\;\text{ft/s}\)
د
\(\text{impact velocity}=-12\sqrt{30}\;\text{ft/s}\)
🚩 إبلاغ
A diver drops from 120 ft above the water. What is the diver's velocity at impact? Ignore air resistance.
أ
\(\text{impact velocity}=-32\sqrt{15}\;\text{ft/s}\)
ب
\(\text{impact velocity}=-16\sqrt{30}\;\text{ft/s}\)
ج
\(\text{impact velocity}=-16\sqrt{3}\;\text{ft/s}\)
د
\(\text{impact velocity}=-12\sqrt{30}\;\text{ft/s}\)
🚩 إبلاغ
If height is increased by a factor of h, by what factor does the impact velocity increase?
أ
\(\sqrt{2h}\)
ب
2h
ج
\(\sqrt{h}\)
د
\(\sqrt{2h}\)
🚩 إبلاغ
\(\text{An object is launched at angle } \theta=\frac{\pi}{3} \text{ from the horizontal with initial speed } v_0=98\;\text{m/s}.\; \text{Determine the time of flight and horizontal range}.\)
أ
\(t=10\sqrt{3}\;\text{sec and } x=490\;\text{m}\)
ب
\(t=10\sqrt{3}\;\text{sec and } x=490\sqrt{3}\;\text{m}\)
ج
\(t=10\sqrt{2}\;\text{sec and } x=490\sqrt{3}\;\text{m}\)
د
\(t=10\sqrt{3}\;\text{sec and } x=490\sqrt{2}\;\text{m}\)
🚩 إبلاغ
Find the time of flight and horizontal range of an object launched at angle 30° with initial speed 40 m/s.
أ
t≈4.8 sec and x≈166.3 m
ب
t≈4.08 sec and x≈144.3 m
ج
t≈4.08 sec and x≈141.3 m
د
t≈4.08 sec and x≈241.3 m
🚩 إبلاغ
\(\text{Evaluate } \int x\sin(4x)\,dx.\)
أ
\(-\frac{1}{16}x\cos(4x)+\frac{1}{4}\sin(4x)+C\)
ب
\(\frac{1}{4}x\cos(4x)-\frac{1}{16}\sin(4x)+C\)
ج
\(-\frac{1}{4}x\cos(4x)+\frac{1}{16}\sin(4x)+C\)
د
\(\frac{1}{4}x\cos(4x)+\frac{1}{16}\sin(4x)+C\)
🚩 إبلاغ
\(\text{Evaluate } \int x e^{2x}\,dx.\)
أ
\(\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+C\)
ب
\(\frac{1}{2}xe^{2x}+\frac{1}{4}e^{2x}+C\)
ج
\(\frac{1}{2}xe^{2x}-\frac{1}{4}e^x+C\)
د
\(\frac{1}{2}xe^{2x}-\frac{1}{4}x^2+C\)
🚩 إبلاغ
\(\text{Evaluate } \int x\ln(x)\,dx.\)
أ
\(\frac{1}{2}x^2\ln(x)+\frac{1}{4}x^2+C\)
ب
\(\frac{1}{2}x^2\ln(x)-\frac{1}{4}x^2+C\)
ج
2x2 ln(x)-4x2 +C
د
\(\frac{1}{2}x\ln(x^2)-\frac{1}{4}x^2+C\)
🚩 إبلاغ
\(\text{Evaluate } \int_{0}^{1}x^2\cos(\pi x)\,dx.\)
أ
\(-\frac{1}{\pi^2}\)
ب
\(\pi^2\)
ج
\(-\frac{2}{\pi^2}\)
د
\(2\pi\)
🚩 إبلاغ
\(\text{Evaluate } \int_{1}^{2}x\ln(x)\,dx.\)
أ
\(2\ln2-\frac{1}{4}\)
ب
\(3\ln2-\frac{3}{4}\)
ج
\(2\ln2-\frac{3}{4}\)
د
\(2\ln2-\frac{1}{2}\)
🚩 إبلاغ
\(\text{Evaluate } \int_{0}^{\pi/4}\cos(2x)\sin^3(2x)\,dx.\)
أ
\(\frac{1}{8}\)
ب
8
ج
-1/8
د
-8
🚩 إبلاغ
\(\text{Evaluate } \int_{\pi/4}^{\pi/3}\cos^3(3x)\sin^3(3x)\,dx.\)
أ
\(\frac{1}{3}\)
ب
1
ج
\(-\frac{1}{72}\)
د
\(\frac{2}{3}\)
🚩 إبلاغ
\(\text{Evaluate } \int_{-\pi/2}^{0}\cos^3(x)\sin(x)\,dx.\)
أ
\(\frac{2}{3}\)
ب
1
ج
\(-\frac{1}{4}\)
د
-1
🚩 إبلاغ
\(\text{Determine } a \text{ if } \int \tan x\,\sec^a(x)\,dx=\frac{1}{3}\sec^3(x)+C.\)
🚩 إبلاغ
\(\text{Evaluate } \int \frac{x^2}{\sqrt{16-x^2}}\,dx.\)
أ
\(8\sin^{-1}\left(\frac{x}{2}\right)-\frac{1}{2}x\sqrt{16-x^2}+C\)
ب
\(8\sin^{-1}\left(\frac{x}{4}\right)-\frac{1}{2}x\sqrt{16-x^2}+C\)
ج
\(\sin^{-1}\left(\frac{x}{4}\right)-\frac{1}{2}x\sqrt{16-x^2}+C\)
د
\(8\sin^{-1}\left(\frac{x}{4}\right)-\frac{1}{4}x\sqrt{16-x^2}+C\)
🚩 إبلاغ
\(\text{Evaluate } \int_{0}^{2}\sqrt{4-x^2}\,dx.\)
أ
\(\pi\)
ب
\(2\pi\)
ج
\(4\pi\)
د
\(\frac{\pi}{2}\)
🚩 إبلاغ
\(\text{Evaluate } \int \frac{x^2}{\sqrt{x^2-9}}\,dx \text{ using trigonometric substitution}.\)
أ
\(\int 9\sec^2\theta\,d\theta\)
ب
\(\int 9\sec^3\theta\,d\theta\)
ج
\(\int 9\sec^2\theta\tan\theta\,d\theta\)
د
\(\int \sec^3\theta\,d\theta\)
🚩 إبلاغ
\(\text{Evaluate } \int x^3\sqrt{x^2-1}\,dx.\)
أ
\(\frac{1}{5}(x^2-1)^{5/2}+\frac{1}{3}(x^2-1)^{3/2}+C\)
ب
\(\frac{1}{5}(x^2-1)^{3/2}+\frac{1}{3}(x^2-1)^{1/2}+C\)
ج
\(\frac{1}{5}(x^3-1)^{5/2}+\frac{1}{3}(x^2-1)^{3/2}+C\)
د
\(\frac{1}{2}(x^2-1)^{5/2}+\frac{1}{3}(x^2-1)^{3/2}+C\)
🚩 إبلاغ
\(\text{Evaluate } \int \frac{2}{\sqrt{x^2-4}}\,dx.\)
أ
\(2\ln\left|x+\sqrt{x^2-4}\right|+C\)
ب
\(2\ln\left|2x+\sqrt{x^2-4}\right|+C\)
ج
\(\ln\left|x+\sqrt{x^2-4}\right|+C\)
د
\(2\ln\left|x-\sqrt{x^2-4}\right|+C\)
🚩 إبلاغ
\(\text{Evaluate } \int \frac{x}{\sqrt{x^2-4}}\,dx.\)
أ
\(\ln\left|\sqrt{x^2-4}\right|+C\)
ب
\(2\sqrt{x^2-4}\)
ج
\(\sqrt{x^2-4}+C\)
د
\(\frac{1}{2\sqrt{x^2-4}}+C\)
🚩 إبلاغ
\(\text{Use partial fractions to find an antiderivative of } \frac{x-5}{x^2-1}.\)
أ
\(3\ln|x+1|-2\ln|x-1|+C\)
ب
\(2\ln|x+1|-2\ln|x-1|+C\)
ج
\(3\ln|x+1|+2\ln|x-1|+C\)
د
\(3\ln|x+1|-2\ln|x+1|+C\)
🚩 إبلاغ
\(\text{Use partial fractions to find an antiderivative of } \frac{6x}{x^2-x-2}.\)
أ
\(3\ln|x+2|-2\ln|x-1|+C\)
ب
\(2\ln|x+1|-2\ln|x-1|+C\)
ج
\(3\ln|x+2|+2\ln|x-2|+C\)
د
\(4\ln|x-2|+2\ln|x+1|+C\)
🚩 إبلاغ
\(\text{Use partial fractions decomposition for } \frac{4x^2+2}{(x^2+1)^2}.\)
أ
\(\frac{4}{x^2+1}+\frac{2}{(x^2+1)^2}\)
ب
\(\frac{4}{x^2+1}-\frac{2}{(x^2+1)^2}\)
ج
\(\frac{2}{x^2+1}-\frac{4}{(x^2+1)^2}\)
د
\(\frac{1}{x^2+1}-\frac{2}{(x^2+1)^2}\)
🚩 إبلاغ
\(\text{Find the form of partial fractions decomposition for } \frac{4x^2+3}{(x^2+x+1)^2}.\)
أ
\(\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}\)
ب
\(\frac{A}{x^2+x+1}+\frac{B}{(x^2+x+1)^2}\)
ج
\(\frac{A}{x^2+x+1}+\frac{Bx+C}{(x^2+x+1)^2}\)
د
\(\frac{Ax+B}{x^2+x+1}+\frac{C}{(x^2+x+1)^2}\)
🚩 إبلاغ
\(\text{Find the solution of } y'=4y \text{ satisfying } y(0)=2.\)
أ
y(t)=-2e3t
ب
y(t)=5e-3t
ج
y(t)=-6e-2t
د
y(t)=2e4t
🚩 إبلاغ
\(\text{Find the solution of } y'=-2y \text{ satisfying } y(0)=-6.\)
أ
y(t)=-2e3t
ب
y(t)=5e-3t
ج
y(t)=-6e-2t
د
y(t)=2e4t
🚩 إبلاغ
\(\text{Find the solution of } y'=2y \text{ satisfying } y(1)=2.\)
أ
y(t)=2et+1
ب
\(y(t)=-\frac{2}{e^2}e^{2t}\)
ج
\(y(t)=\frac{2}{e^2}e^{2t}\)
د
y(t)=2e1-t
🚩 إبلاغ
\(\text{Solve the differential equation } y'=\frac{\cos x}{\sin y}.\)
أ
cos y=-sin x+C
ب
cos x=-sin y+C
ج
cos y=sin x+C
د
sin y=-sin x+C
🚩 إبلاغ
\(\text{Solve the differential equation } y'=x\cos^2(y).\)
أ
\(y=\tan^{-1}\left(\frac{x^2}{2}+C\right)\)
ب
\(y=\tan^{-1}\left(\frac{x^2}{2}\right)+C\)
ج
y=tan-1 (x2 +C)
د
\(y=\tan\left(\frac{x^2}{2}+C\right)\)
🚩 إبلاغ
\(\text{Solve the differential equation } y'=\frac{xy}{1+x^2}.\)
أ
\(y=\sqrt{1+x^2}+C\)
ب
\(y=2k\sqrt{1+x^2}\)
ج
\(y=k\sqrt{1+x^2}\)
د
\(y=k\sqrt{1+x}\)
🚩 إبلاغ
Solve the differential equation y'=2/(xy+y).
أ
\(y=±4ln|1+x^2|+c\)
ب
\(y=±√(2ln|1+x|)+c\)
ج
\(y=±√(4ln|1+x|+c)\)
د
\(y=±√(\ln|1+x^2|)+c\)
