Part B: 10-Mark Questions
QB. 1: (i) By using integration by parts, solve \( \int (\log x)^2 \, dx \)
$$
\begin{aligned}
\text{Sol: Let } I &= \int (\log x)^2 \, dx \\
\text{Put } u &= (\log x)^2 \implies du = 2(\log x) \cdot \frac{1}{x} \, dx \\
\text{Put } dv &= dx \implies v = x \\
\text{Using } \int u \, dv &= uv - \int v \, du: \\
I &= x(\log x)^2 - \int x \cdot \frac{2 \log x}{x} \, dx \\
I &= x(\log x)^2 - 2 \int \log x \, dx \\
\text{Now, solve } \int \log x \, dx & \text{ using parts again (take } u=\log x, dv=dx): \\
\int \log x \, dx &= x \log x - \int x \cdot \frac{1}{x} dx = x \log x - x \\
\text{Substitute back:} \\
I &= x(\log x)^2 - 2 [ x \log x - x ] + C \\
I &= x(\log x)^2 - 2x \log x + 2x + C
\end{aligned}
$$
(ii) Find \( \int_{0}^{1} \tan^{-1}x \, dx \)
$$
\begin{aligned}
\text{Let } u &= \tan^{-1}x, \quad dv = dx \\
du &= \frac{1}{1+x^2} dx, \quad v = x \\
\int_{0}^{1} \tan^{-1}x \, dx &= \left[ x \tan^{-1}x \right]_{0}^{1} - \int_{0}^{1} \frac{x}{1+x^2} \, dx \\
\text{Upper limit: } & 1 \cdot \tan^{-1}(1) = 1 \cdot \frac{\pi}{4} = \frac{\pi}{4} \\
\text{Lower limit: } & 0 \cdot \tan^{-1}(0) = 0 \\
\text{Integral part: } & \text{Multiply and divide by 2 for substitution } (t=1+x^2) \\
&= \frac{\pi}{4} - \frac{1}{2} \int_{0}^{1} \frac{2x}{1+x^2} \, dx \\
&= \frac{\pi}{4} - \frac{1}{2} \left[ \log(1+x^2) \right]_{0}^{1} \\
&= \frac{\pi}{4} - \frac{1}{2} (\log 2 - \log 1) \quad (\text{Since } \log 1 = 0) \\
&= \frac{\pi}{4} - \frac{1}{2} \log 2
\end{aligned}
$$
QB. 2: (i) Evaluate \( \int e^{ax} \sin bx \, dx \) by using integration by parts.
$$
\begin{aligned}
\text{Let } I &= \int e^{ax} \sin bx \, dx \\
\text{Step 1: } u &= e^{ax}, \quad dv = \sin bx \, dx \\
du &= a e^{ax} dx, \quad v = -\frac{\cos bx}{b} \\
I &= -\frac{e^{ax} \cos bx}{b} - \int \left( -\frac{\cos bx}{b} \right) a e^{ax} \, dx \\
I &= -\frac{e^{ax} \cos bx}{b} + \frac{a}{b} \int e^{ax} \cos bx \, dx \\
\text{Step 2: Apply parts again on integral term.} \\
u &= e^{ax}, \quad dv = \cos bx \, dx \implies v = \frac{\sin bx}{b} \\
I &= -\frac{e^{ax} \cos bx}{b} + \frac{a}{b} \left[ \frac{e^{ax} \sin bx}{b} - \int \frac{\sin bx}{b} a e^{ax} dx \right] \\
I &= -\frac{e^{ax} \cos bx}{b} + \frac{a}{b^2} e^{ax} \sin bx - \frac{a^2}{b^2} \int e^{ax} \sin bx \, dx \\
\text{Substitute } I \text{ back into the equation:} \\
I &= -\frac{e^{ax} \cos bx}{b} + \frac{a}{b^2} e^{ax} \sin bx - \frac{a^2}{b^2} I \\
I + \frac{a^2}{b^2} I &= e^{ax} \left( \frac{a \sin bx}{b^2} - \frac{\cos bx}{b} \right) \\
I \left( \frac{b^2 + a^2}{b^2} \right) &= \frac{e^{ax}}{b^2} (a \sin bx - b \cos bx) \\
I &= \frac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx)
\end{aligned}
$$
(ii) Change the order of integration and hence evaluate \( \int_{0}^{4a} \int_{x^2/4a}^{2\sqrt{ax}} xy \, dy \, dx \)
1. Identify the Region R:
Old limits (Vertical Strips): \( y \) goes from \( \frac{x^2}{4a} \) (Bottom) to \( 2\sqrt{ax} \) (Top).
Curves: \( x^2 = 4ay \) (Parabola opening Up) and \( y^2 = 4ax \) (Parabola opening Right).
Intersection points: \( (0,0) \) and \( (4a, 4a) \).
2. Change Order (Horizontal Strips \( dx \, dy \)):
Outer limits (y): \( 0 \) to \( 4a \)
Inner limits (x): Look Left to Right.
- Left Boundary: \( y^2 = 4ax \implies x = \frac{y^2}{4a} \)
- Right Boundary: \( x^2 = 4ay \implies x = \sqrt{4ay} = 2\sqrt{ay} \)
$$
\begin{aligned}
I &= \int_{0}^{4a} \int_{y^2/4a}^{2\sqrt{ay}} xy \, dx \, dy \\
\text{Inner Integral (w.r.t x):} \\
&= y \left[ \frac{x^2}{2} \right]_{y^2/4a}^{2\sqrt{ay}} \\
&= \frac{y}{2} \left[ (2\sqrt{ay})^2 - \left(\frac{y^2}{4a}\right)^2 \right] \\
&= \frac{y}{2} \left[ 4ay - \frac{y^4}{16a^2} \right] = 2ay^2 - \frac{y^5}{32a^2} \\
\text{Outer Integral (w.r.t y):} \\
I &= \int_{0}^{4a} \left( 2ay^2 - \frac{y^5}{32a^2} \right) dy \\
&= \left[ \frac{2ay^3}{3} - \frac{y^6}{192a^2} \right]_{0}^{4a} \\
&= \frac{2a(64a^3)}{3} - \frac{4096a^6}{192a^2} \\
&= \frac{128a^4}{3} - \frac{64a^4}{3} = \frac{64a^4}{3}
\end{aligned}
$$
QB. 3: Establish a reduction formula for \( I_n = \int \sin^n x \, dx \). Hence find \( I_0 \) and \( I_1 \).
$$
\begin{aligned}
I_n &= \int \sin^{n-1}x \sin x \, dx \\
u &= \sin^{n-1}x, \quad dv = \sin x \, dx \implies v = -\cos x \\
I_n &= -\sin^{n-1}x \cos x - \int (-\cos x) \cdot (n-1)\sin^{n-2}x \cos x \, dx \\
I_n &= -\sin^{n-1}x \cos x + (n-1) \int \sin^{n-2}x \cos^2 x \, dx \\
\text{Use } \cos^2 x &= 1 - \sin^2 x: \\
I_n &= -\sin^{n-1}x \cos x + (n-1) \int \sin^{n-2}x (1-\sin^2 x) \, dx \\
I_n &= -\sin^{n-1}x \cos x + (n-1) I_{n-2} - (n-1) I_n \\
I_n + (n-1)I_n &= -\sin^{n-1}x \cos x + (n-1) I_{n-2} \\
n I_n &= -\sin^{n-1}x \cos x + (n-1) I_{n-2} \\
I_n &= \frac{-\sin^{n-1}x \cos x}{n} + \frac{n-1}{n} I_{n-2}
\end{aligned}
$$
Values for \( I_0, I_1 \):
$$
\begin{aligned}
I_0 &= \int \sin^0 x \, dx = \int 1 \, dx = x \\
I_1 &= \int \sin x \, dx = -\cos x
\end{aligned}
$$
QB. 4: (i) Find the area of the region R enclosed by \( y=x^2 \) and \( y=x+2 \).
$$
\begin{aligned}
\text{Intersection: } & x^2 = x+2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1)=0 \\
\text{Limits: } & x \in [-1, 2], \quad y \text{ (vertical strip) } \in [x^2, x+2] \\
\text{Area } A &= \int_{-1}^{2} \int_{x^2}^{x+2} dy \, dx \\
&= \int_{-1}^{2} [y]_{x^2}^{x+2} \, dx = \int_{-1}^{2} (x+2 - x^2) \, dx \\
&= \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2} \\
\text{Upper Limit (2): } & \left( \frac{4}{2} + 4 - \frac{8}{3} \right) = 6 - \frac{8}{3} = \frac{18-8}{3} = \frac{10}{3} \\
\text{Lower Limit (-1): } & \left( \frac{1}{2} - 2 - \frac{-1}{3} \right) = \frac{1}{2} - 2 + \frac{1}{3} = \frac{3-12+2}{6} = -\frac{7}{6} \\
A &= \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} \\
A &= \frac{27}{6} = \frac{9}{2} \text{ sq units.}
\end{aligned}
$$
(ii) Evaluate \( \int_{0}^{3} \int_{0}^{2} e^{x+y} \, dy \, dx \)
$$
\begin{aligned}
I &= \int_{0}^{3} \int_{0}^{2} e^x \cdot e^y \, dy \, dx \\
\text{Since variables are separable:} \\
I &= \left( \int_{0}^{3} e^x dx \right) \cdot \left( \int_{0}^{2} e^y dy \right) \\
&= [e^x]_0^3 \cdot [e^y]_0^2 \\
&= (e^3 - e^0)(e^2 - e^0) \\
&= (e^3 - 1)(e^2 - 1)
\end{aligned}
$$
QB. 5: Evaluate \( \iint xy(x+y) \, dx \, dy \) over the area between \( y=x \) and \( y=x^2 \)
$$
\begin{aligned}
\text{Region: } & x \text{ goes from } 0 \text{ to } 1. \quad y \text{ goes from } x^2 \text{ (parabola) to } x \text{ (line)}. \\
I &= \int_{0}^{1} \int_{x^2}^{x} (x^2y + xy^2) \, dy \, dx \\
\text{Inner } (dy): & \left[ \frac{x^2 y^2}{2} + \frac{x y^3}{3} \right]_{x^2}^{x} \\
&= \left( \frac{x^2(x^2)}{2} + \frac{x(x^3)}{3} \right) - \left( \frac{x^2(x^4)}{2} + \frac{x(x^6)}{3} \right) \\
&= \left( \frac{x^4}{2} + \frac{x^4}{3} \right) - \left( \frac{x^6}{2} + \frac{x^7}{3} \right) = \frac{5x^4}{6} - \frac{x^6}{2} - \frac{x^7}{3} \\
\text{Outer } (dx): & \int_{0}^{1} \left( \frac{5x^4}{6} - \frac{x^6}{2} - \frac{x^7}{3} \right) dx \\
&= \left[ \frac{5}{6}\frac{x^5}{5} - \frac{1}{2}\frac{x^7}{7} - \frac{1}{3}\frac{x^8}{8} \right]_{0}^{1} \\
&= \left[ \frac{x^5}{6} - \frac{x^7}{14} - \frac{x^8}{24} \right]_{0}^{1} \\
&= \frac{1}{6} - \frac{1}{14} - \frac{1}{24} \quad (\text{LCM of 6, 14, 24 is 168}) \\
&= \frac{28 - 12 - 7}{168} = \frac{9}{168} = \frac{3}{56}
\end{aligned}
$$
QB. 6: (i) Using double integral, find the area bounded by \( y=x \) and \( y=x^2 \)
$$
\begin{aligned}
\text{Area } A &= \int_{0}^{1} \int_{x^2}^{x} dy \, dx \\
&= \int_{0}^{1} [y]_{x^2}^{x} \, dx \\
&= \int_{0}^{1} (x - x^2) \, dx \\
&= \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} \\
&= \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \text{ sq units.}
\end{aligned}
$$
(ii) Evaluate \( \int_{0}^{5} \int_{0}^{x^2} x(x^2+y^2) \, dy \, dx \)
$$
\begin{aligned}
I &= \int_{0}^{5} \int_{0}^{x^2} (x^3 + xy^2) \, dy \, dx \\
\text{Inner } (dy): & \left[ x^3 y + \frac{xy^3}{3} \right]_{0}^{x^2} \\
&= x^3(x^2) + \frac{x(x^2)^3}{3} - 0 \\
&= x^5 + \frac{x^7}{3} \\
\text{Outer } (dx): & \int_{0}^{5} \left( x^5 + \frac{x^7}{3} \right) dx \\
&= \left[ \frac{x^6}{6} + \frac{1}{3}\frac{x^8}{8} \right]_{0}^{5} \\
&= \frac{5^6}{6} + \frac{5^8}{24}
\end{aligned}
$$
QB. 7: (i) Evaluate \( \int t^2 e^t \, dt \)
$$
\begin{aligned}
\text{Using Bernoulli's Formula (Tabular Integration):} \\
\int u \, dv &= uv_1 - u'v_2 + u''v_3 - \dots \\
\text{Let } u &= t^2 \text{ (differentiate), } dv = e^t dt \text{ (integrate)} \\
& \begin{array}{c|c}
\text{Derivs (u)} & \text{Integrals (v)} \\
\hline
t^2 & e^t \\
2t & e^t \\
2 & e^t \\
0 & e^t
\end{array} \\
\text{Multiply diagonals with alternating signs } (+, -, +): \\
\text{Result} &= t^2(e^t) - 2t(e^t) + 2(e^t) + C \\
&= e^t(t^2 - 2t + 2) + C
\end{aligned}
$$
(ii) Evaluate \( \int \sin^{-1}x \, dx \)
$$
\begin{aligned}
\text{Let } u &= \sin^{-1}x, \quad dv = dx \\
du &= \frac{1}{\sqrt{1-x^2}} dx, \quad v = x \\
I &= x \sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}} \, dx \\
\text{Sub } t &= 1-x^2 \implies dt = -2x dx \implies x dx = -dt/2 \\
I &= x \sin^{-1}x - \int \frac{-dt/2}{\sqrt{t}} \\
&= x \sin^{-1}x + \frac{1}{2} \int t^{-1/2} dt \\
&= x \sin^{-1}x + \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} + C \\
&= x \sin^{-1}x + \sqrt{1-x^2} + C
\end{aligned}
$$
QB. 8: Find the volume of the tetrahedron bounded by the co-ordinate planes and \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
$$
\begin{aligned}
V &= \int_{0}^{a} \int_{0}^{b(1-\frac{x}{a})} \int_{0}^{c(1-\frac{x}{a}-\frac{y}{b})} dz \, dy \, dx \\
\text{Inner } (dz): &= [z]_0^{\dots} = c\left(1 - \frac{x}{a} - \frac{y}{b}\right) \\
\text{Middle } (dy): &= c \int_{0}^{b(1-x/a)} \left[ \left(1-\frac{x}{a}\right) - \frac{y}{b} \right] dy \\
\text{Let } K &= 1 - x/a. \text{ Limit is } bK. \text{ Integrand is } c(K - y/b). \\
&= c \left[ Ky - \frac{y^2}{2b} \right]_{0}^{bK} \\
&= c \left[ K(bK) - \frac{(bK)^2}{2b} \right] = c \left[ bK^2 - \frac{b^2K^2}{2b} \right] \\
&= c \left[ bK^2 - \frac{bK^2}{2} \right] = \frac{bcK^2}{2} = \frac{bc}{2} \left( 1 - \frac{x}{a} \right)^2 \\
\text{Outer } (dx): &= \frac{bc}{2} \int_{0}^{a} \left( 1 - \frac{x}{a} \right)^2 dx \\
&= \frac{bc}{2} \int_{0}^{a} \left( 1 - \frac{2x}{a} + \frac{x^2}{a^2} \right) dx \\
&= \frac{bc}{2} \left[ x - \frac{2x^2}{2a} + \frac{x^3}{3a^2} \right]_{0}^{a} \\
&= \frac{bc}{2} \left[ a - \frac{a^2}{a} + \frac{a^3}{3a^2} \right] \\
&= \frac{bc}{2} \left[ a - a + \frac{a}{3} \right] = \frac{bc}{2} \cdot \frac{a}{3} = \frac{abc}{6}
\end{aligned}
$$
QB. 9 (i): Evaluate \( \int_{0}^{2a} \int_{0}^{x} \int_{y}^{x} xyz \, dz \, dy \, dx \)
$$
\begin{aligned}
\text{Inner } (dz): & \left[ \frac{xyz^2}{2} \right]_{y}^{x} = \frac{xy}{2} (x^2 - y^2) = \frac{x^3 y - xy^3}{2} \\
\text{Middle } (dy): & \frac{1}{2} \int_{0}^{x} (x^3 y - xy^3) dy \\
&= \frac{1}{2} \left[ \frac{x^3 y^2}{2} - \frac{xy^4}{4} \right]_{0}^{x} \\
&= \frac{1}{2} \left( \frac{x^3(x^2)}{2} - \frac{x(x^4)}{4} \right) = \frac{1}{2} \left( \frac{x^5}{2} - \frac{x^5}{4} \right) \\
&= \frac{1}{2} \left( \frac{2x^5 - x^5}{4} \right) = \frac{x^5}{8} \\
\text{Outer } (dx): & \int_{0}^{2a} \frac{x^5}{8} dx = \frac{1}{8} \left[ \frac{x^6}{6} \right]_{0}^{2a} \\
&= \frac{(2a)^6}{48} = \frac{64a^6}{48} = \frac{4a^6}{3}
\end{aligned}
$$
QB. 9 (ii): Find the volume bounded by the cylinder \( x^2+y^2=4 \) and the planes \( y+z=4 \) and \( z=0 \).
$$
\begin{aligned}
V &= \iint_R z \, dA = \iint_R (4-y) \, dA \\
\text{Polar Coords: } & x = r \cos \theta, y = r \sin \theta, dA = r \, dr \, d\theta \\
\text{Limits: } & \text{Circle radius 2 } \implies r \in [0, 2], \quad \theta \in [0, 2\pi] \\
V &= \int_{0}^{2\pi} \int_{0}^{2} (4 - r \sin \theta) r \, dr \, d\theta \\
&= \int_{0}^{2\pi} \int_{0}^{2} (4r - r^2 \sin \theta) dr \, d\theta \\
\text{Inner } (dr): &= \left[ 2r^2 - \frac{r^3}{3} \sin \theta \right]_{0}^{2} \\
&= 2(4) - \frac{8}{3} \sin \theta = 8 - \frac{8}{3} \sin \theta \\
\text{Outer } (d\theta): &= \int_{0}^{2\pi} \left( 8 - \frac{8}{3} \sin \theta \right) d\theta \\
&= \left[ 8\theta + \frac{8}{3} \cos \theta \right]_{0}^{2\pi} \\
\text{Upper: } & 16\pi + \frac{8}{3} \cos(2\pi) = 16\pi + \frac{8}{3} \\
\text{Lower: } & 0 + \frac{8}{3} \cos(0) = \frac{8}{3} \\
V &= \left( 16\pi + \frac{8}{3} \right) - \frac{8}{3} = 16\pi \text{ cubic units.}
\end{aligned}
$$
QB. 10: Obtain the volume of sphere \( x^2+y^2+z^2=a^2 \) by using triple integral.
$$
\begin{aligned}
\text{Consider 1st octant and multiply by 8:} \\
V &= 8 \int_{0}^{a} \int_{0}^{\sqrt{a^2-x^2}} \int_{0}^{\sqrt{a^2-x^2-y^2}} dz \, dy \, dx \\
\text{Inner } (dz): & [z]_0^{\sqrt{...}} = \sqrt{a^2-x^2-y^2} \\
\text{Middle } (dy): & \int_{0}^{\sqrt{a^2-x^2}} \sqrt{(a^2-x^2)-y^2} \, dy \\
\text{Formula used: } & \int \sqrt{A^2-y^2} dy = \text{Area of circle sector} = \frac{y}{2}\sqrt{A^2-y^2} + \frac{A^2}{2}\sin^{-1}\frac{y}{A} \\
\text{Limits } 0 \to A: & \text{ This is the area of a quarter circle } = \frac{\pi A^2}{4}. \\
\text{Here } A^2 = a^2-x^2, & \text{ so result is } \frac{\pi}{4}(a^2-x^2). \\
\text{Outer } (dx): & 8 \int_{0}^{a} \frac{\pi}{4} (a^2-x^2) dx \\
&= 2\pi \left[ a^2 x - \frac{x^3}{3} \right]_{0}^{a} \\
&= 2\pi \left( a^3 - \frac{a^3}{3} \right) = 2\pi \left( \frac{2a^3}{3} \right) \\
&= \frac{4}{3} \pi a^3 \text{ cubic units.}
\end{aligned}
$$