Anand Classes offers detailed NCERT Solutions for Integrals Exercise 7.8 of Chapter 7 for Class 12 Mathematics, providing students with clear explanations and step-by-step methods to master complex integration problems. These Set-1 notes strictly follow the latest NCERT guidelines and are designed to support effective revision, concept clarity, and strong exam preparation. Click the print button to download study material and notes.
NCERT Question 1: Evaluate the integral
$$\int_{-1}^{1}(x+1)\;dx$$
Solution
Let
$$I=\int_{-1}^{1}(x+1)\;dx$$
First find the antiderivative:
$$\int(x+1)\;dx=\frac{x^{2}}{2}+x$$
Apply the Second Fundamental Theorem of Calculus:
$$I=\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}$$
Substitute the limits:
$$I=\left(\frac{1^{2}}{2}+1\right)-\left(\frac{(-1)^{2}}{2}-1\right)$$
Simplify:
$$I=\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)$$
$$I=\frac{1}{2}+1-\frac{1}{2}+1$$
$$I=2$$
Final Answer
$$\boxed{2}$$
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NCERT Question 2: Evaluate the integral
$$\int_{2}^{3}\frac{1}{x}\;dx$$
Solution
Let
$$I=\int_{2}^{3}\frac{1}{x}\;dx$$
The antiderivative is
$$\int\frac{1}{x}\;dx=\log\lvert x\rvert$$
Apply the Second Fundamental Theorem of Calculus:
$$I=\left[\log\lvert x\rvert\right]_{2}^{3}$$
Substitute the limits:
$$I=\log 3-\log 2$$
Using log properties:
$$I=\log\left(\frac{3}{2}\right)$$
Final Answer
$$\boxed{\log\left(\frac{3}{2}\right)}$$
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NCERT Question 3: Evaluate the integral
$$\int_{1}^{2}\left(4x^{3}-5x^{2}+6x+9\right)\;dx$$
Solution
Let
$$I=\int_{1}^{2}\left(4x^{3}-5x^{2}+6x+9\right)\;dx$$
Find the antiderivative:
$$\int\left(4x^{3}-5x^{2}+6x+9\right)\;dx
= x^{4}-\frac{5x^{3}}{3}+3x^{2}+9x$$
Thus,
$$I=\left[x^{4}-\frac{5x^{3}}{3}+3x^{2}+9x\right]_{1}^{2}$$
Substitute upper limit:
$$2^{4}-\frac{5(2^{3})}{3}+3(2^{2})+9(2)
=16-\frac{40}{3}+12+18$$
Substitute lower limit:
$$1^{4}-\frac{5(1^{3})}{3}+3(1^{2})+9(1)
=1-\frac{5}{3}+3+9$$
Now compute the difference:
$$I=\left(16-\frac{40}{3}+12+18\right)-\left(1-\frac{5}{3}+3+9\right)$$
Simplify:
$$I=33-\frac{35}{3}$$
Write with a common denominator:
$$I=\frac{99-35}{3}$$
$$I=\frac{64}{3}$$
Final Answer
$$\boxed{\frac{64}{3}}$$
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NCERT Question 4: Evaluate the integral
$$\int_{0}^{\dfrac{\pi}{4}} \sin 2x\;dx$$
Solution
Let
$$I=\int_{0}^{\dfrac{\pi}{4}}\sin 2x\;dx$$
Find the antiderivative:
$$\int \sin 2x\;dx = -\frac{\cos 2x}{2}$$
Thus,
$$I=\left[-\frac{\cos 2x}{2}\right]_{0}^{\dfrac{\pi}{4}}$$
Substitute the limits:
$$I=-\frac{1}{2}\left[\cos\left(2\cdot\frac{\pi}{4}\right)-\cos(0)\right]$$
Simplify inside the bracket:
$$I=-\frac{1}{2}\left[\cos\left(\frac{\pi}{2}\right)-\cos(0)\right]$$
Since
$$\cos\left(\frac{\pi}{2}\right)=0 \quad\text{and}\quad \cos(0)=1,$$
we get
$$I=-\frac{1}{2}(0-1)=\frac{1}{2}$$
Final Answer
$$\boxed{\frac{1}{2}}$$
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NCERT Question 5: Evaluate the integral
$$\displaystyle \int_{0}^{\dfrac{\pi}{2}} \cos 2x\;dx$$
Solution
Antiderivative:
$$\int \cos 2x\;dx = \frac{1}{2}\sin 2x$$
Evaluate from $0$ to $\dfrac{\pi}{2}$:
$$\displaystyle I=\left[\frac{1}{2}\sin 2x\right]_{0}^{\dfrac{\pi}{2}}
=\frac{1}{2}\bigl(\sin\pi-\sin 0\bigr)$$
Since $\sin\pi=0$ and $\sin 0=0$,
$$I=\frac{1}{2}(0-0)=0$$
Final Answer
$$\boxed{0}$$
NCERT Question 6: Evaluate the integral
$$\displaystyle \int_{4}^{5} e^{x}\;dx$$
Solution
Antiderivative:
$$\int e^{x}\;dx = e^{x}$$
Evaluate from $4$ to $5$:
$$\displaystyle I=\bigl[e^{x}\bigr]_{4}^{5}=e^{5}-e^{4} = e^{4}(e-1)$$
Final Answer
$$\boxed{e^{5}-e^{4}=e^{4}(e-1)}$$
NCERT Question 7: Evaluate the integral
$$\displaystyle \int_{0}^{\dfrac{\pi}{4}} \tan x\;dx$$
Solution
Antiderivative:
$$\int \tan x\;dx = -\ln\lvert\cos x\rvert$$
Evaluate from $0$ to $\dfrac{\pi}{4}$:
$$\displaystyle I=\left[-\ln\lvert\cos x\rvert\right]_{0}^{\frac{\pi}{4}}
=-\ln\bigl(\cos\frac{\pi}{4}\bigr)+\ln\bigl(\cos 0\bigr)$$
Now $\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$ and $\cos 0=1$, so
$$I=-\ln\left(\frac{1}{\sqrt{2}}\right)+\ln 1
=-\bigl(-\frac{1}{2}\ln 2\bigr)+0=\frac{1}{2}\ln 2$$
Final Answer
$$\boxed{\frac{1}{2}\ln 2}$$
NCERT Question 8: Evaluate the integral
$$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x\;dx$$
Solution
Antiderivative:
$$\int \csc x\;dx = \ln!\bigl|\csc x-\cot x\bigr| + C$$
Evaluate from $\tfrac{\pi}{6}$ to $\tfrac{\pi}{4}$:
$$\displaystyle I=\left[\ln!\bigl|\csc x-\cot x\bigr|\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}
=\ln!\bigl(\csc\tfrac{\pi}{4}-\cot\tfrac{\pi}{4}\bigr)
-\ln!\bigl(\csc\tfrac{\pi}{6}-\cot\tfrac{\pi}{6}\bigr)$$
Compute values:
$$\csc\tfrac{\pi}{4}=\sqrt{2},\quad \cot\tfrac{\pi}{4}=1\quad\Rightarrow\quad \csc\tfrac{\pi}{4}-\cot\tfrac{\pi}{4}=\sqrt{2}-1$$
$$\csc\tfrac{\pi}{6}=2,\quad \cot\tfrac{\pi}{6}=\sqrt{3}\quad\Rightarrow\quad \csc\tfrac{\pi}{6}-\cot\tfrac{\pi}{6}=2-\sqrt{3}$$
So
$$\displaystyle I=\ln\frac{\sqrt{2}-1}{2-\sqrt{3}}$$
(you may leave it as above or combine into a single logarithm)
Final Answer
$$\boxed{\displaystyle \ln\frac{\sqrt{2}-1}{2-\sqrt{3}}}$$
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NCERT Question 9: Evaluate the integral
$$\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}$$
Solution
An antiderivative of the integrand is the inverse sine function:
$$\int \frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}x + C.$$
Apply the limits $0$ to $1$:
$$\displaystyle I=\left[\sin^{-1}x\right]_{0}^{1}=\sin^{-1}(1)-\sin^{-1}(0).$$
Evaluate the values:
$$\sin^{-1}(1)=\frac{\pi}{2},\qquad \sin^{-1}(0)=0.$$
So
$$\displaystyle I=\frac{\pi}{2}-0=\frac{\pi}{2}.$$
Final Answer
$$\boxed{\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}=\frac{\pi}{2}}$$
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NCERT Question 10: Evaluate the integral
$$\displaystyle \int_{0}^{1}\frac{dx}{1+x^{2}}$$
Solution
An antiderivative is
$$\int\frac{dx}{1+x^{2}}=\tan^{-1}x + C.$$
Apply the limits:
$$\left[\tan^{-1}x\right]_{0}^{1}=\tan^{-1}(1)-\tan^{-1}(0)=\frac{\pi}{4}-0=\frac{\pi}{4}.$$
Final Answer
$$\boxed{\displaystyle \frac{\pi}{4}}$$
NCERT Question 11: Evaluate the integral
$$\displaystyle \int_{2}^{3}\frac{dx}{x^{2}-1}$$
Solution
An antiderivative is
$$\int\frac{dx}{x^{2}-1}=\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C.$$
Evaluate from $2$ to $3$:
$$\frac{1}{2}\ln\left|\frac{3-1}{3+1}\right|
-\frac{1}{2}\ln\left|\frac{2-1}{2+1}\right|
=\frac{1}{2}\ln\frac{2/4}{1/3}
=\frac{1}{2}\ln\frac{3}{2}.$$
Final Answer
$$\boxed{\displaystyle \frac{1}{2}\ln\left(\frac{3}{2}\right)}$$
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