1. **Problem Statement:** We will solve multiple integral problems including the fundamental theorem of calculus, average value of functions, definite and indefinite integrals, improper integrals, and partial fractions. --- ### Question 1 **a) Fundamental Theorem of Calculus and definite integral evaluation** 1. The Fundamental Theorem of Calculus states: $$\frac{d}{dx} \int_a^x f(t) dt = f(x)$$ and $$\int_a^b f(t) dt = F(b) - F(a)$$ where $F$ is an antiderivative of $f$. 2. Evaluate $$\int_0^1 (3x^2 + 6) dt$$. Since the integral is with respect to $t$ but the integrand is in terms of $x$, and $x$ is constant with respect to $t$, the integral is: $$\int_0^1 (3x^2 + 6) dt = (3x^2 + 6) \int_0^1 dt = (3x^2 + 6)(1 - 0) = 3x^2 + 6$$ **Answer:** $$3x^2 + 6$$ --- **b) Average value of $f(x) = x^2$ over $[0,6]$ and find $c$ such that $f(c)$ equals average value** 1. Average value formula: $$f_{avg} = \frac{1}{b - a} \int_a^b f(x) dx$$ 2. Compute: $$f_{avg} = \frac{1}{6 - 0} \int_0^6 x^2 dx = \frac{1}{6} \left[ \frac{x^3}{3} \right]_0^6 = \frac{1}{6} \cdot \frac{6^3}{3} = \frac{1}{6} \cdot \frac{216}{3} = \frac{1}{6} \cdot 72 = 12$$ 3. Find $c$ such that $f(c) = c^2 = 12$: $$c = \sqrt{12} = 2\sqrt{3}$$ **Answer:** Average value = 12, $c = 2\sqrt{3}$ --- **c) Evaluate integrals:** **i) $$\int \frac{2x}{1 + x^3} dx$$** 1. Use substitution: let $$u = 1 + x^3$$, then $$du = 3x^2 dx$$. 2. Rewrite integral: $$\int \frac{2x}{u} dx$$ 3. Express $dx$ in terms of $du$: $$du = 3x^2 dx \Rightarrow dx = \frac{du}{3x^2}$$ 4. Substitute back: $$\int \frac{2x}{u} \cdot \frac{du}{3x^2} = \int \frac{2x}{u} \cdot \frac{du}{3x^2} = \int \frac{2}{3u} \cdot \frac{1}{x} du$$ 5. Since $x$ remains, try another approach: rewrite numerator as $2x = \frac{2}{3} \cdot 3x$. 6. Instead, try substitution $w = x^3$, $dw = 3x^2 dx$; not straightforward. 7. Alternatively, rewrite numerator to match derivative of denominator: Derivative of denominator $1 + x^3$ is $3x^2$, numerator is $2x$, not matching. 8. Use substitution $t = x^{3}$, then $dt = 3x^{2} dx$, but numerator is $2x dx$, so no direct substitution. 9. Try substitution $u = x^{2}$, $du = 2x dx$. 10. Then $2x dx = du$, integral becomes: $$\int \frac{du}{1 + (u)^{3/2}}$$ 11. This integral is complicated; leave as is or express answer in terms of $u$. 12. So the integral is: $$\int \frac{2x}{1 + x^3} dx = \int \frac{du}{1 + u^{3/2}}$$ **Answer:** Integral expressed as $$\int \frac{du}{1 + u^{3/2}}$$ where $u = x^2$. **ii) $$\int \cos^2(x) dx$$** 1. Use identity: $$\cos^2(x) = \frac{1 + \cos(2x)}{2}$$ 2. Integral becomes: $$\int \frac{1 + \cos(2x)}{2} dx = \frac{1}{2} \int 1 dx + \frac{1}{2} \int \cos(2x) dx$$ 3. Compute: $$= \frac{x}{2} + \frac{1}{2} \cdot \frac{\sin(2x)}{2} + C = \frac{x}{2} + \frac{\sin(2x)}{4} + C$$ **Answer:** $$\frac{x}{2} + \frac{\sin(2x)}{4} + C$$ **iii) $$\int 3t \cdot e^{t^2} dt$$** 1. Use substitution: $$u = t^2 \Rightarrow du = 2t dt$$ 2. Rewrite integral: $$\int 3t e^{t^2} dt = \int \frac{3}{2} e^u du = \frac{3}{2} \int e^u du = \frac{3}{2} e^u + C = \frac{3}{2} e^{t^2} + C$$ **Answer:** $$\frac{3}{2} e^{t^2} + C$$ **iv) $$\int \frac{x^2 + 3x + 3}{x + 1} dx$$** 1. Perform polynomial division: $$\frac{x^2 + 3x + 3}{x + 1}$$ 2. Divide $x^2$ by $x$ to get $x$. 3. Multiply $x(x + 1) = x^2 + x$. 4. Subtract: $$(x^2 + 3x + 3) - (x^2 + x) = 2x + 3$$ 5. Divide $2x$ by $x$ to get $2$. 6. Multiply $2(x + 1) = 2x + 2$. 7. Subtract: $$(2x + 3) - (2x + 2) = 1$$ 8. So: $$\frac{x^2 + 3x + 3}{x + 1} = x + 2 + \frac{1}{x + 1}$$ 9. Integral becomes: $$\int (x + 2) dx + \int \frac{1}{x + 1} dx = \frac{x^2}{2} + 2x + \ln|x + 1| + C$$ **Answer:** $$\frac{x^2}{2} + 2x + \ln|x + 1| + C$$ --- **d) Evaluate improper integral $$\int_0^2 \frac{1}{\sqrt{4 - 2x}} dx$$ and state convergence** 1. Substitute: $$u = 4 - 2x \Rightarrow du = -2 dx \Rightarrow dx = -\frac{du}{2}$$ 2. Change limits: When $x=0$, $u=4$; when $x=2$, $u=0$. 3. Integral becomes: $$\int_{u=4}^{0} \frac{1}{\sqrt{u}} \cdot \left(-\frac{du}{2}\right) = \frac{1}{2} \int_0^4 u^{-1/2} du$$ 4. Evaluate: $$\frac{1}{2} \left[ 2 u^{1/2} \right]_0^4 = \left[ u^{1/2} \right]_0^4 = 2 - 0 = 2$$ 5. Since the integral evaluates to a finite number, it **converges**. **Answer:** Integral = 2, converges. --- ### Question 2 **a) Evaluate integrals:** **i) $$\int \cos(15t) \cos(11t) dt$$** 1. Use product-to-sum formula: $$\cos A \cos B = \frac{\cos(A-B) + \cos(A+B)}{2}$$ 2. So: $$\int \cos(15t) \cos(11t) dt = \int \frac{\cos(4t) + \cos(26t)}{2} dt = \frac{1}{2} \int \cos(4t) dt + \frac{1}{2} \int \cos(26t) dt$$ 3. Integrate: $$= \frac{1}{2} \cdot \frac{\sin(4t)}{4} + \frac{1}{2} \cdot \frac{\sin(26t)}{26} + C = \frac{\sin(4t)}{8} + \frac{\sin(26t)}{52} + C$$ **Answer:** $$\frac{\sin(4t)}{8} + \frac{\sin(26t)}{52} + C$$ **ii) $$\int (2t^2 + \sqrt{t}) dt$$** 1. Rewrite: $$\int 2t^2 dt + \int t^{1/2} dt = 2 \int t^2 dt + \int t^{1/2} dt$$ 2. Integrate: $$2 \cdot \frac{t^3}{3} + \frac{t^{3/2}}{3/2} + C = \frac{2}{3} t^3 + \frac{2}{3} t^{3/2} + C$$ **Answer:** $$\frac{2}{3} t^3 + \frac{2}{3} t^{3/2} + C$$ **iii) $$\int_0^\pi \frac{\sin^2(x)}{\cos^2(x)} dx$$** 1. Rewrite integrand: $$\frac{\sin^2(x)}{\cos^2(x)} = \tan^2(x)$$ 2. Integral becomes: $$\int_0^\pi \tan^2(x) dx$$ 3. Use identity: $$\tan^2(x) = \sec^2(x) - 1$$ 4. So: $$\int_0^\pi (\sec^2(x) - 1) dx = \int_0^\pi \sec^2(x) dx - \int_0^\pi 1 dx$$ 5. Evaluate: $$[\tan(x)]_0^\pi - [x]_0^\pi = (\tan(\pi) - \tan(0)) - (\pi - 0) = (0 - 0) - \pi = -\pi$$ 6. However, $ an(x)$ is undefined at $x=\frac{\pi}{2}$, so the integral is improper and diverges. **Answer:** Integral diverges due to discontinuity at $x=\frac{\pi}{2}$. --- **b) Area bounded by $y = e^x$, y-axis, and line $x=2$** 1. Area is: $$\int_0^2 e^x dx = [e^x]_0^2 = e^2 - 1$$ **Answer:** $$e^2 - 1$$ --- ### Question 3 **a) Partial fractions and integral of $$f(t) = \frac{12x - 2}{6x^2 - x - 2}$$** 1. Factor denominator: $$6x^2 - x - 2 = (3x + 2)(2x - 1)$$ 2. Express as partial fractions: $$\frac{12x - 2}{(3x + 2)(2x - 1)} = \frac{A}{3x + 2} + \frac{B}{2x - 1}$$ 3. Multiply both sides by denominator: $$12x - 2 = A(2x - 1) + B(3x + 2)$$ 4. Expand: $$12x - 2 = 2Ax - A + 3Bx + 2B = (2A + 3B)x + (-A + 2B)$$ 5. Equate coefficients: $$2A + 3B = 12$$ $$-A + 2B = -2$$ 6. Solve system: From second: $$A = 2B + 2$$ Substitute into first: $$2(2B + 2) + 3B = 12 \Rightarrow 4B + 4 + 3B = 12 \Rightarrow 7B + 4 = 12 \Rightarrow 7B = 8 \Rightarrow B = \frac{8}{7}$$ Then: $$A = 2 \cdot \frac{8}{7} + 2 = \frac{16}{7} + 2 = \frac{16}{7} + \frac{14}{7} = \frac{30}{7}$$ 7. Integral becomes: $$\int \frac{30/7}{3x + 2} dx + \int \frac{8/7}{2x - 1} dx = \frac{30}{7} \int \frac{1}{3x + 2} dx + \frac{8}{7} \int \frac{1}{2x - 1} dx$$ 8. Integrate: $$\frac{30}{7} \cdot \frac{1}{3} \ln|3x + 2| + \frac{8}{7} \cdot \frac{1}{2} \ln|2x - 1| + C = \frac{10}{7} \ln|3x + 2| + \frac{4}{7} \ln|2x - 1| + C$$ **Answer:** $$\frac{10}{7} \ln|3x + 2| + \frac{4}{7} \ln|2x - 1| + C$$ **b) Evaluate $$\int_0^1 3x^2 e^{x^3} dx$$ by substitution** 1. Let: $$u = x^3 \Rightarrow du = 3x^2 dx$$ 2. Change limits: When $x=0$, $u=0$; when $x=1$, $u=1$. 3. Integral becomes: $$\int_0^1 e^u du = [e^u]_0^1 = e - 1$$ **Answer:** $$e - 1$$