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Right choice is (b) 7(e^6 – e^2)

Explanation: \(\int_2^6\)7e^x DX = 7(e^x)^62 dx

= 7(e^6 – e^2)

The correct choice is (c) \(\int_a^b\)f(X)dx+\(\int_b^c\)f(x)dx = \(\int_a^c\)f(x) dx

To explain I WOULD SAY: The ADDING intervals property of definite integrals is \(\int_a^b\)f(x)dx+\(\int_b^c\)f(x)dx.

\(\int_a^b\)f(x)dx+\(\int_b^c\)f(x)dx = \(\int_a^c\)f(x) dx

Right ANSWER is (a) \(\frac{8\sqrt{2}-31}{6}\)

The best I can EXPLAIN: LET \(I=\int_1^2 \sqrt{x}-3x \,dx\)

F(x)=\(\int \sqrt{x}-3x \,dx\)

=\(\frac{x^{1/2+1}}{1/2+1}-\frac{3x^2}{2}=\frac{2x^{\frac{3}{2}}}{3}-\frac{3x^2}{2}\)

By using the second FUNDAMENTAL theorem of calculus, we get

I=F(2)-F(1)=\(\LEFT(\frac{2×2^{3/2}}{3}-\frac{3×2^2}{2}\right)-\left(\frac{2×1^{3/2}}{3}-\frac{3×1^2}{2}\right)\)

I=\(\frac{4\sqrt{2}}{3}-6-\frac{2}{3}+\frac{3}{2}=\frac{8\sqrt{2}-36-4+9}{6}=\frac{8\sqrt{2}-31}{6}\)

The correct choice is (a) \(\FRAC{4}{e}\)

To elaborate: \(I=\int_{-1}^1 \,2xe^X \,dx\)

F(x)=\(\int 2xe^x dx\)

By using the formula, \(\int U.v \,dx=u \int v \,dx-\int u'(\int v \,dx)\)

F(x)=2x\(\int e^x dx-\int(2x)’\int e^x \,dx\)

=\(2xe^x-\int 2e^x dx\)

=\(2e^x (x-1)\)

Therefore, by using the fundamental theorem of calculus, we get

I=F(1)-F(-1)

I=2e^1 (1-1)-2e^-1 (-1-1)

I=\(\frac{4}{e}\).

Right ANSWER is (b) Dummy symbol

For explanation: In \(\int_a^b\)F(y) dy ‘a’ is the called as lower limit and ‘b’ is called the upper limit of the integral. The FUCTION ‘f’ in \(\int_a^b\)f(y) dy is called the integrand. The letter ‘y’ is a dummy symbol and can be REPLACED by any other symbol.

The correct OPTION is (c) \(\int_a^b\)[F(x)-g(x)dx

The explanation is: The SUM difference property of definite INTEGRALS is \(\int_a^b\)[f(x)-g(x)dx

\(\int_a^b\)[f(x)-g(x)dx = \(\int_a^b\)f(x)dx-\(\int_a^b\)g(x)dx

The correct answer is (C) \(\frac{π}{4}\)

To explain I would say: Let \(I=\int_0^{π/2}sin^{2⁡}x \,dx\)

F(x)=\(\INT sin^2⁡x \,dx\)

=\(\int \frac{(1-cos⁡2x)}{2} \,dx\)

=\(\frac{1}{2} (x-\frac{sin⁡2x}{2})\)

APPLYING the limits, we GET

\(I=F(\frac{π}{2})-F(0)=\frac{1}{2} (\frac{π}{2}-\frac{sin⁡π}{2})-\frac{1}{2} (0-\frac{sin⁡0}{2})\)

=\(\frac{1}{4} (π-0)-0=\frac{π}{4}\).

Correct choice is (d) sin (3) – sin (2)

The BEST I can explain: \(\int_2^3\)cosx DX = (sin x)^32

= sin (3) – sin (2)

Correct answer is (d) SIN(x)+\(\FRAC {3}{4}\)x^-3+c

For explanation: ∫cos(x)-\(\frac {3}{X4}\)dx = ∫cos(x) dx−∫3(x^-4) dx

= sin(x)+\(\frac {3}{4}\)x^-3+c

Right answer is (d) –\(\frac{5}{3} \,x \,cos⁡3x+\frac{5}{9} \,sin⁡3x+C\)

Easy EXPLANATION: By using the formula, ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) ,we get

∫ 5x sin⁡3x dx=5x∫ sin⁡3x dx -∫ (5x)’∫ sin⁡3x dx

=\(5x(-\frac{cos⁡3x}{3})+\INT \frac{5 cos⁡3x}{3} dx\)

=-\(\frac{5}{3} x cos⁡3x+\frac{5}{9} sin⁡3x+C\)

Right answer is (a) \(\FRAC{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\)

To EXPLAIN I WOULD SAY: ∫ 10 log⁡x.x^2 dx=10∫ log⁡x.x^2 dx

By using the formula, ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get

\(10\int log⁡x.x^2 \,dx=10(log⁡x \int x^2 \,dx-\int (log⁡x)’\int \,x^2 \,dx)\)

=\(10 \left(\frac{x^3 log⁡x}{3}-(\int \frac{1}{x}.\frac{x^3}{3} \,dx)\right)+C\)

=\(10 \left(\frac{x^3 log⁡x}{3}-\frac{1}{3} \frac{x^3}{3}\right)+C\)

=\(\frac{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\).

Correct OPTION is (d) \(2\SQRT{x^5+9}\)

For explanation I would say: Let x^5+9=t

Differentiating w.r.t x, we get

5x^4 dx=dt

\(\INT \frac{5x^4}{\sqrt{x^5+9}} dx=\int \frac{dt}{\sqrt{t}}\)

=\(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2\sqrt{t}\)

REPLACING t with x^5+9, we get

\(\int \frac{5x^4}{\sqrt{x^5+9}} dx=2\sqrt{x^5+9}\).

The correct answer is (b) \(sin⁡x^3+8x^3+C\)

EASY explanation: By using the METHOD of INTEGRATION by substitution,

Let x^3=t

Differentiating w.r.t x, we GET

3x^2 dx=dt

\(\int 3x^2 \,(cos⁡x^3+8) \,dx=\int (cos⁡t+8)dt\)

\(\int (cos⁡t+8) dt=sin⁡t+8t\)

Replacing t with x^3,we get

\(\int 3x^2 (cos⁡x^3+8) dx=sin⁡x^3+8x^3+C\)

The correct answer is (c) 5

The best explanation: LET \(I=\int_0^{\frac{π}{2}} \,5 \,sin⁡X \,DX\)

F(x)=\(\int5 \,sin⁡x \,dx=-5 \,cos⁡x\)

Applying the limits by using the fundamental THEOREM of calculus, we get

I=F(\(\frac{π}{2}\))-F(0)

∴\(\int_0^{\frac{π}{2}} \,5 \,sin⁡x \,dx=-5[cos⁡\frac{π}{2}-cos⁡0]\)

=-5[0-1]=5

Correct ANSWER is (a) 2

The BEST explanation: \(\int_0^{\PI }\)sin⁡x dx = [- COS x]^x0

= – cos π – (-cos 0)

= -(-1)-(-1)

= 2

Correct OPTION is (a) True

Explanation: \(\int_0^{\pi }\)sin⁡ y dy = [-COS y]^y0

= – cos π – (- cos 0)

= -(-1)-(-1)

= 2

Right OPTION is (c) \(\frac{7}{2} (x^2 log⁡x-x)+C\)

To explain I WOULD say: ∫ 7 log⁡x.x dx=7∫ log⁡x.x dx

Using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx) , we get

7∫ log⁡x.x dx=7(log⁡x ∫ x dx-(log⁡x)’∫ x dx)

=\(7\left (\frac{x^2 log⁡x}{2}-\frac{1}{x}.\frac{x^2}{2}\right)\)

=\(\frac{7}{2} (x^2 log⁡x-x)+C\)

The correct option is (b) \(\frac{e^{2x}}{4} (2x-1)+C\)

Explanation: By using the FORMULA \(\INT u.V \,dx=u\int \,v \,dx-\int u'(\int \,v \,dx)\) we GET

\(\int XE^{2x} dx=x\int e^{2x} dx-\int (x)’\int e^{2x} \,dx\)

=\(\frac{xe^{2x}}{2}-\int 1.\frac{e^{2x}}{2} \,dx\)

=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)

=\(\frac{e^{2x}}{4} (2x-1)+C\)

The correct answer is (d) \(\frac{7}{6} log⁡|\frac{x-3}{x+3}|+C\)

Explanation: \(\int\frac{7dx}{x^2-9}=2\int \frac{7dx}{x^2-3^2}\)

By USING the formula \(\int \frac{dx}{x^2-a^2}=\frac{1}{2A} log⁡|\frac{x-a}{x+a}|+C\)

∴\(7\int \frac{dx}{x^2-3^2}=7(\frac{1}{2(3)} log⁡|\frac{x-3}{x+3}|)+7C_1\)

=\(\frac{7}{6} log⁡|\frac{x-3}{x+3}|+C\)

Right option is (C) –\(\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C\)

Explanation: To FIND: ∫ 3 sin^3⁡(x+2) DX

We know that, sin⁡3x=3 sin⁡x-4 sin^3⁡x

∴sin^3⁡⁡x=\(\frac{3 sin⁡x-sin⁡3x}{4}\)

sin^3⁡(x+2)=\(\frac{(3 sin⁡(x+2)-sin⁡(3x+6))}{4}\)

\(\int sin^3⁡(x+2) \,dx=\frac{3}{4} \int sin⁡(x+2) \,dx-\frac{1}{4} \int \,sin⁡(3x+6) \,dx\)

=-\(\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C\)

Right option is (a) –\(\FRAC{1}{(3E^{x^3})}+C\)

To explain: Let x^3=t

3x^2 dx=dt

x^2 dx=dt/3

∴\(\INT \frac{x^2}{e^{x^3}} dx=\frac{1}{3} \int \frac{dt}{e^t}\)

=\(\frac{1}{3} \left (-e^{-t}\right )\)

Replacing t with x^3, we get

\(\int \frac{x^2}{e^{x^3}} dx=-\frac{1}{3e^{x^3}}+C\)

The correct answer is (d) \(3e^x+\FRAC{1}{3} (log⁡x)^2+C\)

EXPLANATION: \(\INT 3e^x+\frac{2(log⁡x^2)}{3x} dx=3\int e^x dx+\frac{2}{3} \int \frac{log⁡x}{x}\)

Let log⁡x=t

Differentiating w.r.t x, we GET

\(\frac{1}{x} dx=dt\)

∴\(\int \frac{log⁡x}{x}=\int \,t \,dt=\frac{t^2}{2}\)

\(\int e^x dx=e^x\)

Replacing t with log⁡x, we get

\(\int 3e^x+\frac{2(log⁡x^2)}{3x} dx=3e^x+\frac{1}{3} (log⁡x)^2+C\)

Right ANSWER is (c) \(\int_a^b\)f(X)dx = 0

Best EXPLANATION: The zero-length interval property is one of the properties used in definite integrals and they are ALWAYS positive. The zero-length interval property is \(\int_a^b\)f(x)dx = 0.

Correct choice is (c) 4-2\(\sqrt{2}\)

To explain I WOULD SAY: I=\(\int_{-1}^1 \FRAC{5x^4}{\sqrt{x^5+3}} dx\)

Let x^5+3=t

Differentiating w.r.t x, we get

5x^4 dx=dt

The new limits

when x=-1,t=2

when x=1,t=4

∴\(\int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx=\int_2^4 \frac{dt}{\sqrt{t}}\)

=\([2\sqrt{t}]_2^4=2(\sqrt{4}-\sqrt{2})=4-2\sqrt{2}\)

Right option is (c) \(\FRAC {1}{4}\) (SIN 3 + COS 3 – sin 2 – cos 2)

To elaborate: \(\int_2^3 \frac {cos⁡x-sin⁡x}{4}\)dx = \(\frac {1}{4}\) [sin x – (- cos x)]^32

= \(\frac {1}{4}\) (sin x + cos x)^32

= \(\frac {1}{4}\) (sin 3 + cos 3) – \(\frac {1}{4}\) (sin 2 + cos 2)

= \(\frac {1}{4}\) (sin 3 + cos 3 – sin 2 – cos 2)

Correct option is (d) \(\int_a^b\)f(x)dx=-\(\int_b^a\)f(x)dx

Best explanation: In the REVERSE INTEGRAL property the UPPER LIMITS and lower limits are interchanged. The reverse integral property of definite integrals is \(\int_a^b\)f(x)dx=-\(\int_b^a\)f(x)dx.

Right choice is (B) \(\frac{3}{2}\)

Easiest explanation: Let \(I=\int_{-1}^23x+x^2-2 \,dx\)

F(x)=\(\int 3x+x^2-2 \,dx\)

=\(\frac{3x^2}{2}+\frac{x^3}{3}-2x\)

APPLYING the limits, we get

I=F(2)-F(-1)

I=\(\left(\frac{(3×2^3)}{2}+\frac{2^3}{3}-2(2)\right)-\left(\frac{3 (-1)^2}{2}+\frac{(-1)^3}{3}-2(-1)\right)\)

I=\(6+\frac{8}{3}-4-\frac{3}{2}+\frac{1}{3}-2=\frac{3}{2}\).

Right choice is (b) \(\FRAC{1}{\sqrt{2}}\)

The explanation is: I=\(\int_0^{\frac{\sqrt{π}}{2}} \,2X \,cos⁡ x^2 \,dx\)

Let x^2=t

Differentiating w.r.t x, we get

2x dx=DT

The NEW limits

When x=0,t=0

When \(x={\frac{\sqrt{π}}{2}}, t=\frac{π}{4}\)

∴\(\int_0^{\frac{\sqrt{π}}{2}} \,2x \,cos⁡ x^2 \,dx=\int_0^{\frac{π}{4}} \,cos⁡t \,dt\)

\(I =[sin⁡t]_0^{\frac{π}{4}}=sin⁡ \frac{π}{4}-sin⁡0=1/\sqrt{2}\).

Right choice is (c) cos⁡X (log⁡(cos⁡x)-1)+C

To explain I would SAY: LET cos⁡x=t

Differentiating w.r.t x, we get

-sin⁡x dx=dt

sin⁡x dx=-dt

∴∫ sin⁡x log⁡(cos⁡x) dx=∫ -log⁡t dt

Using ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) , we get

∫ -log⁡t dt=-log⁡t ∫ 1 dt-∫ (-log⁡t)’ ∫ 1 dt

=-t log⁡t+∫ dt

=-t log⁡t+t

=t(log⁡t-1)

REPLACING t with cos⁡x, we get

∫ sin⁡x log⁡(cos⁡x) dx=cos⁡x (log⁡(cos⁡x)-1)+C

The correct choice is (b) division

To elaborate: An IMPROPER integration FACTOR can be reduced to proper fraction by division, i.e., if the NUMERATOR and DENOMINATOR have same degree, then they must be DIVIDED in order to reduce it to proper fraction.

Correct option is (d) \(\frac{3}{5}log|x+2| + \frac{1}{5}log|x^2+1|+\frac{1}{5} tan^{-1}x+C\)

Explanation: \(\INT \frac{(x^2+x+1)dx}{(x+2)(x^2+1)} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+1)}\)

Now equating, (x^2+x+1) = A (x^2+1) + (Bx+C) (x+2)

After equating and SOLVING for coefficient we get values,

A=\(\frac{3}{5}\), B=\(\frac{2}{5}\), C=\(\frac{1}{5}\), now putting these values in the EQUATION we get,

\(\int \frac{(x^2+x+1)dx}{(x+2)(x^2+1)} = \frac{3}{5} \int \frac{dx}{(x+2)} + \frac{1}{5} \int \frac{2xdx}{(x^2+1)} + \frac{1}{5} \int \frac{dx}{(x^2+1)}\)

Hence it COMES, \(\frac{3}{5} log|x+2| + \frac{1}{5} log|x^2+1|+\frac{1}{5}tan^{-1}x+C\)

Right answer is (c) \(\FRAC{1}{2} TAN^{-1}⁡\frac{x-4}{2}+C\)

Explanation: \(\int \frac{dx}{x^2-8x+20}=\int \frac{dx}{(x^2-2(4x)+4^2)+4}\)

=\(\int \frac{dx}{(x-4)^2+2^2}\)

Let x-4=t

Differentiating w.r.t x, we GET

dx=dt

By using the formula \(\int \frac{dx}{x^2+a^2}=\frac{1}{a} tan^{-1}⁡\frac{x}{a}+C\)

\(\int \frac{dx}{(x-4)^2+2^2}=\int \frac{dt}{t^2+2^2}=\frac{1}{2} tan^{-1}⁡\frac{t}{2}+C\)

Replacing t with x-4, we get

\(\int \frac{dx}{(x-4)^2+2^2}=\frac{1}{2} tan^{-1}⁡\frac{x-4}{2}+C\)

The CORRECT CHOICE is (a) log⁡2

To explain: \(I=\int_0^{π/4} \,2 \,tan⁡x \,dx\)

F(x)=∫ 2 tan⁡x dx

=2∫ tan⁡x dx

=2 log⁡|sec⁡x|

Therefore, by USING the fundamental theorem of calculus, we get

I=F(π/4)-F(0)

\(=2\left(log⁡|sec \frac{⁡π}{4}|-log⁡|sec⁡0|\right)=2 log⁡\SQRT{2}-log⁡1\)

\(=2 log⁡\sqrt{2}=log⁡(\sqrt{2})^2=log⁡2\)

I=\(\frac{8}{3} log⁡2-\frac{8}{3}-0+\frac{1}{3}=\frac{8}{3} log⁡2-\frac{7}{3}\).

Right option is (d) \(\frac{A}{(x-a)} + \frac{B}{(x-b)}\)

Easiest explanation: The given FUNCTION \(\frac{px+q}{(x-a)(x-b)}\), a≠b can also be WRITTEN as

\(\frac{A}{(x-a)} + \frac{B}{(x-b)}\) and is further used to solve INTEGRATION by partial fractions numerical.

The correct answer is (b) –\(\FRAC{3}{2} cos⁡x+\frac{cos⁡3x}{6}+x+C\)

The best explanation: We KNOW that, sin⁡3x=3 sin⁡x-4 sin^3⁡x

∴sin^3x=\(\frac{3 sin⁡x-sin⁡3x}{4}\)

\(\INT 2 \,sin^3⁡x+1 \,dx=\int \frac{(3 sin⁡x-sin⁡3x)}{2} dx+\int dx\)

=\(\frac{3}{2} \int sin⁡x dx-\frac{1}{2} \int sin⁡3x dx+\int dx\)

=-\(\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+x+C\)

Correct option is (a) 2 tan^4⁡x+C

The EXPLANATION: To FIND: \(\int 8 \,tan^3⁡x \,sec^2⁡x \,DX\)

LET tan⁡x=t

sec^2⁡x dx=dt

∴\(\int 8 \,tan^3⁡x \,sec^2⁡x \,dx=\int 8 \,t^3 \,dt=\frac{8t^4}{4}=2t^4\)

REPLACING t with tan⁡x, we get

\(\int 8 tan^3⁡x sec^2⁡x dx=2 tan^4⁡x+C\)

Correct OPTION is (a) \(\frac{7X^4}{4}-2e^{2X}+\frac{2}{x}+C\)

The best I can explain: To find:\(\int 7x^8-4e^{2x}-\frac{2}{x^2} DX\)

\(\int \,7x^8-4e^{2x}-\frac{2}{x^2} \,dx=\int 7x^9 dx-4\int e^{2x} dx-2\int \frac{1}{x}^2 dx\)

\(\int \,7x^8-4e^{2x}-\frac{2}{x^2} \,dx=\frac{7x^{9+1}}{9+1}-\frac{4e^{2x}}{2}-\frac{2x^{-2+1}}{-2+1}\)

∴\(\int \,7x^8-4e^{2x}-\frac{2}{x^2} dx=\frac{7x^{10}}{10}-2e^{2x}+\frac{2}{x}+C\)

Correct ANSWER is (c) \(\FRAC{4x^{5/2}}{6}+\frac{2x^{7/2}}{7}+C\)

The BEST explanation: To FIND \(\int (2+x)x\sqrt{x} dx\)

\(\int \,(2+x)x\sqrt{x} \,dx=\int \,2x\sqrt{x}+x^{5/2} \,dx\)

\(\int \,(2+x)x\sqrt{x} \,dx=\int \,2x^{3/2} dx + \int x^{5/2} dx\)

\(\int \,(2+x)x\sqrt{x} \,dx=\frac{2x^{3/2+1}}{3/2+1}+\frac{x^{5/2+1}}{5/2+1}\)

\(\int \,(2+x)x\sqrt{x} \,dx=\frac{4x^{5/2}}{5}+\frac{2x^{7/2}}{7}+C\)

Correct choice is (a) 32

The explanation is: LET I=\(\int_0^8x \,dx\)

F(X)=\(\int x \,dx=\frac{x^2}{2}\)

Using the SECOND fundamental theorem of calculus, we get

I=F(8)-f(0)

∴\(\int_0^8x \,dx=\frac{8^2}{2}-0=32\)